WORK AND KINETIC ENERGY 6.1 IDENTIFY and SET UP: For parts (a) through (d), identify the appropriate value of φ and use the relation W = FP s = ( F cos φ ) s In part (e), apply the relation Wnet = Wstudent + Wgrav + Wn + W f EXECUTE: (a) Since you are applying a horizontal force, φ = 0° Thus, Wstudent = (2.40 N)(cos0°)(1.50 m) = 3.60 J (b) The friction force acts in the horizontal direction, opposite to the motion, so φ = 180° W f = ( F f cos φ ) s = (0.600 N)(cos180°)(1.50 m) = −0.900 J (c) Since the normal force acts upward and perpendicular to the tabletop, φ = 90° Wn = (n cos φ ) s = (ns )(cos90°) = 0.0 J (d) Since gravity acts downward and perpendicular to the tabletop, φ = 270° Wgrav = ( mg cos φ )s = (mgs )(cos 270°) = 0.0 J (e) Wnet = Wstudent + Wgrav + Wn + W f = 3.60 J + 0.0 J + 0.0 J − 0.900 J = 2.70 J 6.2 EVALUATE: Whenever a force acts perpendicular to the direction of motion, its contribution to the net work is zero IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so W = F s cos φ SET UP: In part (a), when the cable pulls horizontally φ = 0° and when it pulls at 35.0° above the horizontal φ = 35.0° In part (b), if the cable pulls horizontally φ = 180° If the cable pulls on the car at 35.0° above the horizontal it pulls on the truck at 35.0° below the horizontal and φ 145.0° For the gravity force φ = 90°, since the force is vertical and the displacement is horizontal EXECUTE: (a) When the cable is horizontal, W = (850 N)(5.00 × 103 m)cos0° = 4.26 × 106 J When the cable is 35.0° above the horizontal, W = (850 N)(5.00 × 103 m)cos35.0° = 3.48 × 106 J (b) cos180° = − cos0° and cos145.0° = − cos35.0°, so the answers are −4.25 × 106 J and −3.48 × 106 J (c) Since cos φ = cos90° = 0, W = in both cases 6.3 EVALUATE: If the car and truck are taken together as the system, the tension in the cable does no net work IDENTIFY: Each force can be used in the relation W = F|| s = ( F cos φ ) s for parts (b) through (d) For part (e), apply the net work relation as Wnet = Wworker + Wgrav + Wn + W f SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, Fworker = f k = μk n EXECUTE: (a) The magnitude of the force the worker must apply is: Fworker = f k = μ k n = μ k mg = (0.25)(30.0 kg)(9.80 m/s ) = 74 N (b) Since the force applied by the worker is horizontal and in the direction of the displacement, φ = 0° and the work is: Wworker = ( Fworker cos φ ) s = [(74 N)(cos0°)](4.5 m) = +333 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-1 6-2 Chapter (c) Friction acts in the direction opposite of motion, thus φ = 180° and the work of friction is: W f = ( f k cos φ ) s = [(74 N)(cos180°)](4.5 m) = −333 J (d) Both gravity and the normal force act perpendicular to the direction of displacement Thus, neither force does any work on the crate and Wgrav = Wn = 0.0 J (e) Substituting into the net work relation, the net work done on the crate is: Wnet = Wworker + Wgrav + Wn + W f = +333 J + 0.0 J + 0.0 J − 333 J = 0.0 J EVALUATE: The net work done on the crate is zero because the two contributing forces, Fworker and F f , 6.4 are equal in magnitude and opposite in direction IDENTIFY: The forces are constant so Eq (6.2) can be used to calculate the work Constant speed implies G G a = We must use ΣF = ma applied to the crate to find the forces acting on it (a) SET UP: The free-body diagram for the crate is given in Figure 6.4 EXECUTE: ΣFy = ma y n − mg − F sin 30° = n = mg + F sin 30° f k = μ k n = μ k mg + F μ k sin 30° Figure 6.4 ΣFx = ma x F cos30° − f k = F cos30° − μ k mg − μ k sin 30° F = μk mg 0.25(30.0 kg)(9.80 m/s ) = = 99.2 N cos30° − μ k sin 30° cos30° − (0.25)sin 30° (b) WF = ( F cos φ ) s = (99.2 N)(cos30°)(4.5 m) = 387 J F= G G ( F cos30° is the horizontal component of F ; the work done by F is the displacement times the G component of F in the direction of the displacement.) (c) We have an expression for f k from part (a): f k = μ k (mg + F sin 30°) = (0.250)[(30.0 kg)(9.80 m/s ) + (99.2 N)(sin 30°)] = 85.9 N φ = 180° since f k is opposite to the displacement Thus W f = ( f k cos φ ) s = (85.9 N)(cos180°)(4.5 m) = −387 J (d) The normal force is perpendicular to the displacement so φ = 90° and Wn = The gravity force (the weight) is perpendicular to the displacement so φ = 90° and Ww = (e) Wtot = WF + W f + Wn + Ww = +387 J + (−387 J) = 6.5 EVALUATE: Forces with a component in the direction of the displacement positive work, forces opposite to the displacement negative work and forces perpendicular to the displacement zero work The total work, obtained as the sum of the work done by each force, equals the work done by the net force In this problem, Fnet = since a = and Wtot = 0, which agrees with the sum calculated in part (e) IDENTIFY: The gravity force is constant and the displacement is along a straight line, so W = Fs cos φ SET UP: The displacement is upward along the ladder and the gravity force is downward, so φ = 180.0° − 30.0° = 150.0° w = mg = 735 N EXECUTE: (a) W = (735 N)(2.75 m)cos150.0° = -1750 J (b) No, the gravity force is independent of the motion of the painter EVALUATE: Gravity is downward and the vertical component of the displacement is upward, so the gravity force does negative work © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6.6 6-3 IDENTIFY and SET UP: WF = ( F cos φ ) s, since the forces are constant We can calculate the total work by summing the work done by each force The forces are sketched in Figure 6.6 EXECUTE: W1 = F1s cos φ1 W1 = (1.80 × 106 N)(0.75 × 103 m)cos14° W1 = 1.31 × 109 J W2 = F2 s cos φ = W1 Figure 6.6 6.7 Wtot = W1 + W2 = 2(1.31 × 109 J) = 2.62 × 109 J EVALUATE: Only the component F cos φ of force in the direction of the displacement does work These G components are in the direction of s so the forces positive work IDENTIFY: All forces are constant and each block moves in a straight line, so W = Fs cosφ The only direction the system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right SET UP: Since the 12.0 N block moves at constant speed, a = for it and the tension T in the string is T = 12.0 N Since the 20.0 N block moves to the right at constant speed the friction force f k on it is to the left and f k = T = 12.0 N EXECUTE: (a) (i) φ = 0° and W = (12.0 N)(0.750 m)cos 0° = 9.00 J (ii) φ = 180° and W = (12.0 N)(0.750 m)cos180° = -9.00 J (b) (i) φ = 90° and W = (ii) φ = 0° and W = (12.0 N)(0.750 m)cos 0° = 9.00 J (iii) φ = 180° and W = (12.0 N)(0.750 m)cos180° = −9.00 J (iv) φ = 90° and W = (c) Wtot = for each block 6.8 6.9 EVALUATE: For each block there are two forces that work, and for each block the two forces work of equal magnitude and opposite sign When the force and displacement are in opposite directions, the work done is negative IDENTIFY: Apply Eq (6.5) SET UP: iˆ ⋅ iˆ = ˆj ⋅ ˆj = and iˆ ⋅ ˆj = ˆj ⋅ iˆ = G G EXECUTE: The work you is F ⋅ s = ((30 N) iˆ − (40 N) ˆj ) ⋅ ((−9.0 m)iˆ − (3.0 m) ˆj ) G G F ⋅ s = (30 N)( −9.0 m) + (−40 N)(−3.0 m) = −270 N ⋅ m + 120 N ⋅ m = −150 J G G EVALUATE: The x-component of F does negative work and the y-component of F does positive work G The total work done by F is the sum of the work done by each of its components IDENTIFY: Apply Eq (6.2) or (6.3) G G SET UP: The gravity force is in the − y -direction, so Fmg ⋅ s = -mg ( y2 − y1 ) EXECUTE: (a) (i) Tension force is always perpendicular to the displacement and does no work (ii) Work done by gravity is − mg ( y2 − y1 ) When y1 = y2 , Wmg = (b) (i) Tension does no work (ii) Let l be the length of the string Wmg = − mg ( y2 − y1 ) = − mg (2l ) = −25.1 J EVALUATE: In part (b) the displacement is upward and the gravity force is downward, so the gravity force does negative work © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-4 6.10 Chapter IDENTIFY and SET UP: Use W = Fp s = ( F cosφ ) s to calculate the work done in each of parts (a) through (c) In part (d), the net work consists of the contributions due to all three forces, or wnet = wgrav + wn + wf Figure 6.10 EXECUTE: (a) As the package slides, work is done by the frictional force which acts at φ = 180° to the displacement The normal force is mg cos53.0° Thus for μ k = 0.40, W f = Fp s = ( f k cos φ ) s = ( μ k n cos φ ) s = [ μk (mg cos53.0°)](cos180°) s W f = (0.40)[(8.00 kg)(9.80 m/s )(cos53.0°)](cos180°)(2.00 m) = −38 J (b) Work is done by the component of the gravitational force parallel to the displacement φ = 90° − 53° = 37° and the work of gravity is Wgrav = (mg cosφ ) s = [(8.00 kg)(9.80 m/s )(cos37.0°)](2.00 m) = + 125 J (c) Wn = since the normal force is perpendicular to the displacement (d) The net work done on the package is Wnet = Wgrav + Wn + W f = 125 J + 0.0 J − 38 J = 87 J 6.11 EVALUATE: The net work is positive because gravity does more positive work than the magnitude of the negative work done by friction IDENTIFY: Since the speed is constant, the acceleration and the net force on the monitor are zero SET UP: Use the fact that the net force on the monitor is zero to develop expressions for the friction force, f k , and the normal force, n Then use W = FP s = ( F cosφ ) s to calculate W Figure 6.11 EXECUTE: (a) Summing forces along the incline, ΣF = ma = = f k − mg sin θ , giving f k = mg cosθ , directed up the incline Substituting gives W f = ( f k cosφ ) s = [( mg sinθ )cosφ ]s W f = [(10.0 kg)(9.80 m/s )(sin36.9°)](cos0°)(5.50 m) = +324 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6-5 (b) The gravity force is downward and the displacement is directed up the incline so φ = 126.9° Wgrav = (10.0 kg)(9.80 m/s )(cos 126.9°)(5.50 m) = −324 J 6.12 (c) The normal force, n, is perpendicular to the displacement and thus does zero work EVALUATE: Friction does positive work and gravity does negative work The net work done is zero IDENTIFY: We want to find the work done by a known force acting through a known displacement G G G SET UP: W = F ⋅ s = Fx sx + Fy s y We know the components of F but need to find the components of the G displacement s G EXECUTE: Using the magnitude and direction of s , its components are x = (48.0 m)cos 240.0o = -24.0 m and y = (48.0 m)sin 240.0o = −41.57 m Therefore, G s = (−24.0 m) iˆ + (−41.57 m) ˆj The definition of work gives G G W = F ⋅ s = ( −68.0 N)(−24.0 m) + (36.0 N)( −41.57 m) = +1632 J − 1497 J = +135 J 6.13 EVALUATE: The mass of the car is not needed since it is the given force that is doing the work IDENTIFY: Find the kinetic energy of the cheetah knowing its mass and speed SET UP: Use K = 12 mv to relate v and K EXECUTE: (a) K = 6.14 mv = (70 kg)(32 m/s) = 3.6 × 104 J 2 (b) K is proportional to v , so K increases by a factor of when v doubles EVALUATE: A running person, even with a mass of 70 kg, would have only 1/100 of the cheetah’s kinetic energy since a person’s top speed is only about 1/10 that of the cheetah IDENTIFY: The book changes its speed and hence its kinetic energy, so work must have been done on it SET UP: Use the work-kinetic energy theorem Wnet = K f − Ki , with K = 12 mv In part (a) use Ki and K f to calculate W In parts (b) and (c) use Ki and W to calculate K f EXECUTE: (a) Substituting the notation i = A and f = B, Wnet = K B − K A = 12 (1.50 kg)[(1.25 m/s) − (3.21 m/s)2 ] = − 6.56 J (b) Noting i = B and f = C , KC = K B + Wnet = 12 (1.50 kg)(1.25 m/s)2 − 0.750 J = + 0.422 J KC = 12 mvC2 so vC = KC /m = 0.750 m/s (c) Similarly, KC = 12 (1.50 kg)(1.25 m/s) + 0.750 J = 1.922 J and vC = 1.60 m/s EVALUATE: Negative Wnet corresponds to a decrease in kinetic energy (slowing down) and positive Wnet corresponds to an increase in kinetic energy (speeding up) 6.15 IDENTIFY: K = 12 mv Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic energy SET UP: v = 12 km/s = 1.2 × 104 m/s A 1.0 megaton bomb releases 4.184 × 1015 J of energy EXECUTE: (a) K = 12 (1.4 × 108 kg)(1.2 × 104 m/s) = 1.0 × 1016 J 6.16 1.0 × 1016 J = 2.4 The energy is equivalent to 2.4 one-megaton bombs 4.184 × 1015 J EVALUATE: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large crater IDENTIFY: Use the equations for free-fall to find the speed of the weight when it reaches the ground and use the formula for kinetic energy SET UP: Kinetic energy is K = 12 mv The mass of an electron is 9.11 × 10-31 kg In part (b) take + y (b) downward, so a y = +9.80 m/s and v 2y = v02y + 2a y ( y − y0 ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-6 Chapter EXECUTE: (a) K = 12 (9.11 × 10−31 kg)(2.19 × 106 m/s)2 = 2.18 × 10-18 J (b) v 2y = v02y + 2a y ( y − y0 ) gives v y = 2(9.80 m/s )(1 ⋅ m) = 4.43 m/s K = 12 (1.0 kg)(4.43 m/s) = 9.8 J (c) Solving K = 12 mv for v gives v = 6.17 2K 2(100 J) = = 2.6 m/s Yes, this is reasonable m 30 kg EVALUATE: A running speed of m/s corresponds to running a 100-m dash in about 17 s, so 2.6 m/s is reasonable for a running child IDENTIFY: Newton’s second law applies to the system of blocks, as well as the work-energy theorem SET UP: Newton’s second law is ΣFx = ma x and the work-energy theorem is Wtot = ΔK = K f − Ki EXECUTE: (a) For the hanging block, Newton’s second law gives 12.0 N − T = (1.224 kg) a and for the block on the table T = (2.041 kg)a 12.0 N = (3.265 kg)a This gives a = 3.675 m/s and T = 7.50 N (b) (i) WT = T (1.20 m) = (7.50 N)(1.20 m) = +9.00 J (ii) Wmg = mg (1.20 m) = (12.0 N)(1.20 m) = 14.4 J WT = -T (1.20 m) = -9.00 J Wtot = 5.40 J (c) For the system of two blocks, Wtot = +9.00 J + 5.40 J = 14.4 J This equals the work done by gravity on the 12.0 N block The total work done by T is zero 1 (d) Wtot = ΔK = K f − Ki Since Ki = 0, K f = (2.041 kg)v + (1.224 kg)v 2 ⎡1 ⎤ Therefore 14.4 J = ⎢ (2.041 kg) + (1.224 kg) ⎥ v gives v = 2.97 m/s ⎣2 ⎦ EVALUATE: As a check, we could find the velocity in part (d) using the standard kinematics formulas since the acceleration is constant: v = + 2ax = 2(3.675 m/s )(1.20 m) gives the same answer as in (d) 6.18 IDENTIFY: Only gravity does work on the watermelon, so Wtot = Wgrav Wtot = ΔK and K = 12 mv SET UP: Since the watermelon is dropped from rest, K1 = EXECUTE: (a) Wgrav = mgs = (4.80 kg)(9.80 m/s )(25.0 m) = 1180 J (b) Wtot = K − K1 so K = 1180 J v = 2K2 2(1180 J) = = 22.2 m/s 4.80 kg m (c) The work done by gravity would be the same Air resistance would negative work and Wtot would be less than Wgrav The answer in (a) would be unchanged and both answers in (b) would decrease 6.19 EVALUATE: The gravity force is downward and the displacement is downward, so gravity does positive work IDENTIFY: Wtot = K − K1 In each case calculate Wtot from what we know about the force and the displacement SET UP: The gravity force is mg, downward The friction force is f k = μ k n = μk mg and is directed opposite to the displacement The mass of the object isn’t given, so we expect that it will divide out in the calculation EXECUTE: (a) K1 = Wtot = Wgrav = mgs mgs = 12 mv22 and v2 = gs = 2(9.80 m/s )(95.0 m) = 43.2 m/s (b) K = (at the maximum height) Wtot = Wgrav = -mgs − mgs = - 12 mv12 and v1 = gs = 2(9.80 m/s )(525 m) = 101 m/s (c) K1 = 12 mv12 K = Wtot = W f = -μk mgs − μk mgs = - 12 mv12 s= v12 2μk g = (5.00 m/s) 2(0.220)(9.80 m/s ) = 5.80 m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy (d) K1 = 12 mv12 K = 12 mv22 Wtot = W f = -μk mgs K = Wtot + K1 mv 2 6-7 = -μk mgs + 12 mv12 v2 = v12 − μ k gs = (5.00 m/s) − 2(0.220)(9.80 m/s )(2.90 m) = 3.53 m/s (e) K1 = 12 mv12 K = Wgrav = -mgy2 , where y2 is the vertical height −mgy2 = - 12 mv12 and (12.0 m/s) v12 = = 7.35 m g 2(9.80 m/s ) EVALUATE: In parts (c) and (d), friction does negative work and the kinetic energy is reduced In part (a), gravity does positive work and the speed increases In parts (b) and (e), gravity does negative work and the speed decreases The vertical height in part (e) is independent of the slope angle of the hill IDENTIFY: From the work-energy relation, W = Wgrav = ΔK rock y2 = 6.20 SET UP: As the rock rises, the gravitational force, F = mg , does work on the rock Since this force acts in the direction opposite to the motion and displacement, s, the work is negative Let h be the vertical distance the rock travels EXECUTE: (a) Applying Wgrav = K − K1 we obtain −mgh = 12 mv22 − 12 mv12 Dividing by m and solving for v1, v1 = v22 + gh Substituting h = 15.0 m and v2 = 25.0 m/s, v1 = (25.0 m/s) + 2(9.80 m/s )(15.0 m) = 30.3 m/s (b) Solve the same work-energy relation for h At the maximum height v2 = − mgh = 12 mv22 − 12 mv12 and h = 6.21 v12 − v22 (30.3 m/s) − (0.0 m/s)2 = = 46.8 m 2g 2(9.80 m/s ) EVALUATE: Note that the weight of 20 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have an initial velocity of 30.3 m/s As the rock moves upward gravity does negative work and this reduces the kinetic energy of the rock IDENTIFY and SET UP: Apply Eq (6.6) to the box Let point be at the bottom of the incline and let point be at the skier Work is done by gravity and by friction Solve for K1 and from that obtain the required initial speed EXECUTE: Wtot = K − K1 K1 = 12 mv02 , K = Work is done by gravity and friction, so Wtot = Wmg + W f Wmg = -mg ( y2 − y1 ) = -mgh W f = - fs The normal force is n = mg cos α and s = h/sin α , where s is the distance the box travels along the incline W f = -(μk mg cos α )( h/sin α ) = -μk mgh/tan α Substituting these expressions into the work-energy theorem gives −mgh − μk mgh/tan α = − 12 mv02 Solving for v0 then gives v0 = gh(1 + μk / tan α ) 6.22 EVALUATE: The result is independent of the mass of the box As α → 90°, h = s and v0 = gh , the same as throwing the box straight up into the air For α = 90° the normal force is zero so there is no friction IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK SET UP: Parallel to incline: force component W|| = mg sinα , down incline; displacement s = h/sinα , down incline Perpendicular to the incline: s = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-8 Chapter EXECUTE: (a) W|| = (mg sin α )( h/sin α ) = mgh W|| = 0, since there is no displacement in this direction Wmg = W|| + W⊥ = mgh, same as falling height h (b) Wtot = K − K1 gives mgh = 12 mv and v = gh , same as if had been dropped from height h The 6.23 work done by gravity depends only on the vertical displacement of the object When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller (c) h = 15.0 m, so v = gh = 17.1s EVALUATE: The acceleration and time of travel are different for an object sliding down an incline and an object in free-fall, but the final velocity is the same in these two cases IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK SET UP: φ = 0° EXECUTE: From Eqs (6.1), (6.5) and (6.6), and solving for F, 2 2 1 ΔK m(v2 − v1 ) (8.00 kg)((6.00 m/s) − (4.00 m/s) ) = = = 32.0 N s s (2.50 m) EVALUATE: The force is in the direction of the displacement, so the force does positive work and the kinetic energy of the object increases IDENTIFY and SET UP: Use Eq (6.6) to calculate the work done by the foot on the ball Then use Eq (6.2) to find the distance over which this force acts EXECUTE: Wtot = K − K1 F= 6.24 K1 = 12 mv12 = 12 (0.420 kg)(2.00 m/s) = 0.84 J K = 12 mv22 = 12 (0.420 kg)(6.00 m/s) = 7.56 J Wtot = K − K1 = 7.56 J − 0.84 J = 6.72 J 6.25 The 40.0 N force is the only force doing work on the ball, so it must 6.72 J of work WF = ( F cosφ )s gives that W 6.72 J s= = = 0.168 m F cos φ (40.0 N)(cos0) EVALUATE: The force is in the direction of the motion so positive work is done and this is consistent with an increase in kinetic energy IDENTIFY: Apply Wtot = ΔK SET UP: v1 = 0, v2 = v f k = μk mg and f k does negative work The force F = 36.0 N is in the direction of the motion and does positive work EXECUTE: (a) If there is no work done by friction, the final kinetic energy is the work done by the applied force, and solving for the speed, 2W Fs 2(36.0 N)(1.20 m) = = = 4.48 m/s v= m m (4.30 kg) (b) The net work is Fs − f k s = ( F − μk mg ) s, so v= 6.26 2( F − μk mg ) s 2(36.0 N − (0.30)(4.30 kg)(9.80 m/s )(1.20 m) = = 3.61 m/s m (4.30 kg) EVALUATE: The total work done is larger in the absence of friction and the final speed is larger in that case IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK SET UP: The gravity force has magnitude mg and is directed downward EXECUTE: (a) On the way up, gravity is opposed to the direction of motion, and so W = -mgs = -(0.145 kg)(9.80 m/s )(20.0 m) = -28.4 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy (b) v2 = v12 + 6.27 6-9 W 2(−28.4 J) = (25.0 m/s) + = 15.3 m/s m (0.145 kg) (c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up On the way down, gravity will have done both negative and positive work on the ball, but the net work at this height will be the same EVALUATE: As the baseball moves upward, gravity does negative work and the speed of the baseball decreases (a) IDENTIFY and SET UP: Use Eq (6.2) to find the work done by the positive force Then use Eq (6.6) to find the final kinetic energy, and then K = 12 mv22 gives the final speed EXECUTE: Wtot = K − K1, so K = Wtot + K1 K1 = 12 mv12 = 12 (7.00 kg)(4.00 m/s) = 56.0 J The only force that does work on the wagon is the 10.0 N force This force is in the direction of the displacement so φ = 0° and the force does positive work: WF = ( F cos φ ) s = (10.0 N)(cos0)(3.0 m) = 30.0 J Then K = Wtot + K1 = 30.0 J + 56.0 J = 86.0 J K = 12 mv22 ; v2 = 2K2 2(86.0 J) = = 4.96 m/s m 7.00 kg G G (b) IDENTIFY: Apply ΣF = ma to the wagon to calculate a Then use a constant acceleration equation to calculate the final speed The free-body diagram is given in Figure 6.27 SET UP: EXECUTE: ΣFx = ma x F = ma x ax = F 10.0 N = = 1.43 m/s m 7.00 kg Figure 6.27 v22x = v12x + 2a2 ( x − x0 ) v2 x = v12x + 2a x ( x − x0 ) = (4.00 m/s) + 2(1.43 m/s )(3.0 m) = 4.96 m/s 6.28 EVALUATE: This agrees with the result calculated in part (a) The force in the direction of the motion does positive work and the kinetic energy and speed increase In part (b), the equivalent statement is that the force produces an acceleration in the direction of the velocity and this causes the magnitude of the velocity to increase IDENTIFY: Apply Wtot = K − K1 SET UP: K1 = The normal force does no work The work W done by gravity is W = mgh , where h = L sin θ is the vertical distance the block has dropped when it has traveled a distance L down the incline and θ is the angle the plane makes with the horizontal EXECUTE: The work-energy theorem gives v = 2K 2W = = gh = gL sin θ Using the given m m numbers, v = 2(9.80 m/s )(0.75 m)sin 36.9° = 2.97 m/s EVALUATE: The final speed of the block is the same as if it had been dropped from a height h © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-10 6.29 Chapter IDENTIFY: Wtot = K − K1 Only friction does work SET UP: Wtot = W f k = -μ k mgs K = (car stops) K1 = 12 mv02 EXECUTE: (a) Wtot = K − K1 gives − μ k mgs = - 12 mv02 s = (b) (i) μ kb = μ ka sμ k = 2μk g ⎛μ ⎞ v02 = constant so sa μ ka = sb μ kb sb = ⎜ ka ⎟ sa = sa /2 The minimum stopping 2g ⎝ μ kb ⎠ distance would be halved (ii) v0b = 2v0 a s v02 = 2μk g ⎛v ⎞ sb = sa ⎜ 0b ⎟ = sa 2 v0 a v0b ⎝ v0 a ⎠ sμ = 2v0 a , μ kb = μ ka 2k = = constant, 2g v0 = constant, so The stopping distance would become times as great (iii) v0b sa = sb ⎛ μ ⎞⎛ v ⎞ ⎛1⎞ sb = sa ⎜ ka ⎟⎜ 0b ⎟ = sa ⎜ ⎟ ( ) = sa The stopping distance would double 2 v0 a v0b ⎝2⎠ ⎝ μkb ⎠⎝ v0a ⎠ EVALUATE: The stopping distance is directly proportional to the square of the initial speed and indirectly proportional to the coefficient of kinetic friction IDENTIFY: We know (or can calculate) the change in the kinetic energy of the crate and want to find the work needed to cause this change, so the work-energy theorem applies SET UP: Wtot = ΔK = K f − Ki = 12 mvf2 − 12 mvi2 so 6.30 v02 sa μ ka = sb μ kb EXECUTE: Wtot = K f − Ki = 12 (30.0 kg)(5.62 m/s)2 − 12 (30.0 kg)(3.90 m/s)2 Wtot = 473.8 J − 228.2 J = 246 J 6.31 EVALUATE: Kinetic energy is a scalar and does not depend on direction, so only the initial and final speeds are relevant IDENTIFY: The elastic aortal material behaves like a spring, so we can apply Hooke’s law to it SET UP: Fspr = F , where F is the pull on the strip or the force the strip exerts, and F = kx EXECUTE: (a) Solving F = kx for k gives k = (b) F = kx = (40.0 N/m)(0.0114 m) = 0.456 N 6.32 F 1.50 N = = 40.0 N/m x 0.0375 m EVALUATE: It takes 0.40 N to stretch this material by 1.0 cm, so it is not as stiff as many laboratory springs IDENTIFY: The work that must be done to move the end of a spring from x1 to x2 is W = 12 kx22 − 12 kx12 The force required to hold the end of the spring at displacement x is Fx = kx SET UP: When the spring is at its unstretched length, x = When the spring is stretched, x > 0, and when the spring is compressed, x < 2W 2(12.0 J) EXECUTE: (a) x1 = and W = 12 kx22 k = = = 2.67 × 104 N/m x2 (0.0300 m) (b) Fx = kx = (2.67 × 104 N/m)(0.0300 m) = 801 N (c) x1 = 0, x2 = -0.0400 m W = 12 (2.67 × 104 N/m)(−0.0400 m) = 21.4 J Fx = kx = (2.67 × 104 N/m)(0.0400 m) = 1070 N 6.33 EVALUATE: When a spring, initially unstretched, is either compressed or stretched, positive work is done by the force that moves the end of the spring IDENTIFY: The springs obey Hook’s law and balance the downward force of gravity SET UP: Use coordinates with + y upward Label the masses 1, 2, and and call the amounts the springs are stretched x1, x2 , and x3 Each spring force is kx © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-20 Chapter EXECUTE: (a) WF = ( F cos φ ) s G G Both F and s are parallel to the incline and in the same direction, so φ = 90° and WF = Fs = (140 N)(3.80 m) = 532 J (b) The directions of the displacement and of the gravity force are shown in Figure 6.63b Ww = ( w cos φ ) s φ = 115°, so Ww = (196 N)(cos115°)(3.80 m) Ww = −315 J Figure 6.63b Alternatively, the component of w parallel to the incline is w sin 25° This component is down the incline G so its angle with s is φ = 180° Ww sin 25° = (196 Nsin 25°)(cos180°)(3.80 m) = −315 J The other G component of w, w cos 25°, is perpendicular to s and hence does no work Thus Ww = Ww sin 25° = −315 J, which agrees with the above (c) The normal force is perpendicular to the displacement (φ = 90°), so Wn = (d) n = w cos 25° so f k = μ k n = μ k w cos 25° = (0.30)(196 N)cos 25° = 53.3 N W f = ( f k cos φ ) x = (53.3 N)(cos180°)(3.80 m) = −202 J (e) Wtot = WF + Ww + Wn + W f = +532 J − 315 J + − 202 J = 15 J (f) Wtot = K − K1, K1 = 0, so K = Wtot mv22 6.64 = Wtot so v2 = 2Wtot 2(15 J) = = 1.2 m/s m 20.0 kg EVALUATE: The total work done is positive and the kinetic energy of the suitcase increases as it moves up the incline IDENTIFY: The work he does to lift his body a distance h is W = mgh The work per unit mass is (W/m) = gh SET UP: The quantity gh has units of N/kg EXECUTE: (a) The man does work, (9.8 N/kg)(0.4 m) = 3.92 J/kg (b) (3.92 J/kg)/(70 J/kg) × 100 = 5.6% (c) The child does work (9.8 N/kg)(0.2 m) = 1.96 J/kg (1.96 J/kg)/(70 J/kg) × 100 = 2.8% (d) If both the man and the child can work at the rate of 70 J/kg, and if the child only needs to use 1.96 J/kg instead of 3.92 J/kg, the child should be able to more chin-ups 6.65 EVALUATE: Since the child has arms half the length of his father’s arms, the child must lift his body only 0.20 m to a chin-up IDENTIFY: Four forces act on the crate: the 290-N push, gravity, friction, and the normal force due to the surface of the ramp The total work is the sum of the work due to all four of these forces The acceleration is constant because the forces are constant SET UP: The work is W = Fs cos φ We can use the standard kinematics formulas because the acceleration is constant The work-energy theorem, Wtot = ΔK , applies EXECUTE: (a) First calculate the work done by each of the four forces The normal force does no work because it is perpendicular to the displacement The other work is WF = ( F cos34.0o )(15.0 m) = (290 N)(cos34.0o )(15.0 m) = 3606 J, Wmg = − mg (15.0 m)(sin 34.0o ) = −(20.0 kg)(9.8 m/s2 )(15.0 m)(sin 34.0o ) = −1644 J and W f = − f (15.0 m) = −(65.0 N)(15.0 m) = −975 J The total work is Wtot = 3606 J − 1644 J − 975 J = 987 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy (b) First find the final velocity: Wtot = ΔK so vf = 6-21 2(987 J) = 9.935 m/s 20.0 kg ⎛ v + vx ⎞ Constant acceleration gives x − x0 = ⎜ x ⎟ t so ⎝ ⎠ t= 6.66 2( x − x0 ) 2(15.0 m) = = 3.02 s v0 x + vx + (9.935 m/s) EVALUATE: Work is a scalar, so we can algebraically add the work done by each of the forces G G IDENTIFY: Apply ΣF = ma to each block to find the tension in the string Each force is constant and W = Fs cos φ 20.0 N = 2.04 kg and SET UP: The free-body diagram for each block is given in Figure 6.66 m A = g mB = 12.0 N = 1.22 kg g EXECUTE: T − f k = m Aa wB − T = mB a wB − f k = (m A + mB ) a ⎛ wB ⎞ ⎛ mA ⎞ ⎛ wA ⎞ (a) f k = a = ⎜ and T = wB ⎜ ⎟ = wB ⎜ ⎟ = 7.50 N + m m ⎝ m A + mB ⎟⎠ B⎠ ⎝ A ⎝ wA + wB ⎠ 20.0 N block: Wtot = Ts = (7.50 N)(0.750 m) = 5.62 J 12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 7.50 N)(0.750 m) = 3.38 J (b) f k = μ k wA = 6.50 N a = wB − μ k wA m A + mB ⎛ mA ⎞ ⎛ wA ⎞ T = f k + ( wB − μk wA ) ⎜ ⎟ = μk wA + ( wB − μk wA ) ⎜ ⎟ + m m B⎠ ⎝ A ⎝ wA + wB ⎠ T = 6.50 N + (5.50 N)(0.625) = 9.94 N 20.0 N block: Wtot = (T − f k ) s = (9.94 N − 6.50 N)(0.750 m) = 2.58 J 12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 9.94 N)(0.750 m) = 1.54 J EVALUATE: Since the two blocks move with equal speeds, for each block Wtot = K − K1 is proportional to the mass (or weight) of that block With friction the gain in kinetic energy is less, so the total work on each block is less Figure 6.66 6.67 IDENTIFY: K = 12 mv Find the speed of the shuttle relative to the earth and relative to the satellite SET UP: Velocity is distance divided by time For one orbit the shuttle travels a distance 2π R EXECUTE: (a) ⎛ 2π R ⎞ mv = m ⎜ ⎟ = (86,400 kg) 2 ⎝ T ⎠ ⎛ 2π (6.66 × 106 m) ⎜⎜ ⎝ (90.1 min)(60 s/min) ⎞ 12 ⎟⎟ = 2.59 × 10 J ⎠ (b) (1/2) mv = (1/2)(86,400 kg)((1.00 m)/(3.00 s))2 = 4.80 × 103 J EVALUATE: The kinetic energy of an object depends on the reference frame in which it is measured © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-22 6.68 Chapter IDENTIFY: W = Fs cos φ and Wtot = K − K1 SET UP: f k = μk n The normal force is n = mg cosθ , with θ = 24.0° The component of the weight parallel to the incline is mg sin θ EXECUTE: (a) φ = 180° and W f = − f k s = −( μk mg cosθ ) s = −(0.31)(5.00 kg)(9.80 m/s2 )(cos 24.0°)(1.50 m) = −20.8 J (b) (5.00 kg)(9.80 m/s )(sin24.0°)(1.50 m) = 29.9 J (c) The normal force does no work (d) Wtot = 29.9 J − 20.8 J = +9.1 J (e) K = K1 + Wtot = (1/2)(5.00 kg)(2.2 m/s)2 + 9.1 J = 21.2 J, and so v2 = 2(21.2 J)/(5.00 kg) = 2.9 m/s 6.69 EVALUATE: Friction does negative work and gravity does positive work The net work is positive and the kinetic energy of the object increases IDENTIFY: The initial kinetic energy of the head is absorbed by the neck bones during a sudden stop Newton’s second law applies to the passengers as well as to their heads SET UP: In part (a), the initial kinetic energy of the head is absorbed by the neck bones, so 12 mvmax = 8.0 J For part (b), assume constant acceleration and use vf = vi + at with vi = 0, to calculate a; then apply Fnet = ma to find the net accelerating force Solve: (a) vmax = 2(8.0 J) = 1.8 m/s = 4.0 mph 5.0 kg vf − vi 1.8 m/s − = = 180 m/s ≈ 18 g , and Fnet = ma = (5.0 kg)(180 m/s ) = 900 N −3 t 10.0 × 10 s EVALUATE: The acceleration is very large, but if it lasts for only 10 ms it does not much damage IDENTIFY: The force does work on the object, which changes its kinetic energy, so the work-energy theorem applies The force is variable so we must integrate to calculate the work it does on the object (b) a = 6.70 x2 F ( x) dx x1 SET UP: Wtot = ΔK = K f − Ki = 12 mvf2 − 12 mvi2 and Wtot = ∫ EXECUTE: Wtot = ∫ x2 x1 F ( x)dx = ∫ 5.00 m [ −12.0 N + (0.300 N/m ) x ]dx Wtot = −(12.0 N)(5.00 m) + (0.100 N/m )(5.00 m)3 = −60.0 J + 12.5 J = −47.5 J Wtot = 12 mvf2 − 12 mvi2 = −47.5 J, so the final velocity is vf = vi2 − 6.71 2(47.5 J) 2(47.5 J) = (6.00 m/s) − = 4.12 m/s m 5.00 kg EVALUATE: We could not readily this problem by integrating the acceleration over time because we know the force as a function of x, not of t The work-energy theorem provides a much simpler method IDENTIFY: Apply Eq (6.7) dx SET UP: ∫ = − x x x ⎛1 1⎞ dx ⎡ 1⎤ k = − − ⎢ x ⎥ = k ⎜ x − x ⎟ The force is given to be attractive, x1 x1 x ⎣ ⎦ x1 1⎠ ⎝ 1 < , and W < so Fx < , and k must be positive If x2 > x1 , x2 x1 EXECUTE: x2 (a) W = ∫ Fx dx = −k ∫ x2 (b) Taking “slowly” to be constant speed, the net force on the object is zero The force applied by the hand ⎛1 1⎞ is opposite Fx , and the work done is negative of that found in part (a), or k ⎜ − ⎟ , which is positive if ⎝ x1 x2 ⎠ x2 > x1 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6.72 6-23 (c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero EVALUATE: Your force is directed away from the origin, so when the object moves away from the origin your force does positive work IDENTIFY: Apply Eq (6.6) to the motion of the asteroid SET UP: Let point be at a great distance and let point be at the surface of the earth Assume K1 = From the information given about the gravitational force its magnitude as a function of distance r from the center of the earth must be F = mg ( RE /r ) This force is directed in the − rˆ direction since it is a “pull.” F is not constant so Eq (6.7) must be used to calculate the work it does 2 RE ⎛ mgR ⎞ EXECUTE: W = − ∫ F ds = − ∫ ⎜ E ⎟ dr = − mgRE2 (−(1/r ) ∞RE ) = mgRE ∞ ⎝ r ⎠ Wtot = K − K1 , K1 = This gives K = mgRE = 1.25 × 1012 J K = 12 mv22 so v2 = K /m = 11,000 m/s EVALUATE: 6.73 IDENTIFY: Note that v2 = gRE ; the impact speed is independent of the mass of the asteroid Calculate the work done by friction and apply Wtot = K − K1 Since the friction force is not constant, use Eq (6.7) to calculate the work SET UP: Let x be the distance past P Since μ k increases linearly with x, μ k = 0.100 + Ax When x = 12.5 m, μ k = 0.600, so A = 0.500/(12.5 m) = 0.0400/m (a) Wtot = ΔK = K − K1 gives − ∫ μk mgdx = − mv12 Using the above expression for μk , 2 x2 ⎡ ⎤ 1 x g ∫ (0.100 + Ax)dx = v12 and g ⎢(0.100) x2 + A ⎥ = v12 2 ⎣ ⎦ EXECUTE: ⎡ x2 ⎤ (9.80 m/s ) ⎢(0.100) x2 + (0.0400/m) ⎥ = (4.50 m/s) Solving for x2 gives x2 = 5.11 m 2⎦ ⎣ (b) μ k = 0.100 + (0.0400/m)(5.11 m) = 0.304 v2 (4.50 m/s) (c) Wtot = K − K1 gives − μk mgx2 = − mv12 x2 = = = 10.3 m μk g 2(0.100)(9.80 m/s ) 6.74 EVALUATE: The box goes farther when the friction coefficient doesn’t increase IDENTIFY: Use Eq (6.7) to calculate W SET UP: x1 = In part (a), x2 = 0.050 m In part (b), x2 = −0.050 m x2 x2 0 EXECUTE: (a) W = ∫ Fdx = ∫ (kx − bx + cx3 ) dx = k b c x2 − x2 + x2 W = (50.0 N/m) x22 − (233 N/m ) x23 + (3000 N/m3 ) x24 When x2 = 0.050 m, W = 0.12 J (b) When x2 = −0.050 m, W = 0.17 J (c) It’s easier to stretch the spring; the quadratic −bx term is always in the − x -direction, and so the needed force, and hence the needed work, will be less when x2 > 6.75 EVALUATE: When x = 0.050 m, Fx = 4.75 N When x = −0.050 m, Fx = −8.25 N G G IDENTIFY and SET UP: Use ΣF = ma to find the tension force T The block moves in uniform circular G G motion and a = arad (a) The free-body diagram for the block is given in Figure 6.75 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-24 Chapter EXECUTE: ΣFx = max T =m v2 R T = (0.0900 kg) (0.70 m/s) = 0.11 N 0.40 m Figure 6.75 v2 (2.80 m/s) = (0.0900 kg) = 7.1 N R 0.10 m (c) SET UP: The tension changes as the distance of the block from the hole changes We could use (b) T = m x2 F x1 x W =∫ dx to calculate the work But a much simpler approach is to use Wtot = K − K1 EXECUTE: The only force doing work on the block is the tension in the cord, so Wtot = WT K1 = 12 mv12 = 12 (0.0900 kg)(0.70 m/s) = 0.0221 J, K = 12 mv22 = 12 (0.0900 kg)(2.80 m/s) = 0.353 J, so Wtot = K − K1 = 0.353 J − 0.0221 J = 0.33 J This is the amount of work done by the person who pulled 6.76 the cord EVALUATE: The block moves inward, in the direction of the tension, so T does positive work and the kinetic energy increases IDENTIFY: Use Eq (6.7) to find the work done by F Then apply Wtot = K − K1 SET UP: dx ∫ x2 = − x x2 x1 EXECUTE: W = ∫ ⎛1 ⎞ dx = α ⎜ − ⎟ x ⎝ x1 x2 ⎠ α W = (2.12 × 10−26 N ⋅ m )((0.200 m −1) − (1.25 × 109 m −1 )) = −2.65 × 10−17 J Note that x1 is so large compared to x2 that the term 1/x1 is negligible Then, using Eq (6.13) and solving for v2 , v2 = v12 + 2W 2( −2.65 × 10−17 J) = (3.00 × 105 m/s) + = 2.41 × 105 m/s m (1.67 × 10−27 kg) (b) With K = 0, W = − K1 Using W = − 2α , 2(2.12 × 10−26 N ⋅ m ) = 2.82 × 10−10 m mv12 (1.67 × 10−27 kg)(3.00 × 105 m/s) (c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00 × 105 m/s EVALUATE: As the proton moves toward the uranium nucleus the repulsive force does negative work and the kinetic energy of the proton decreases As the proton moves away from the uranium nucleus the repulsive force does positive work and the kinetic energy of the proton increases G G G G IDENTIFY and SET UP: Use vx = dx/dt and ax = dvx /dt Use ΣF = ma to calculate F from a x2 = 6.77 α α x2 K1 = = EXECUTE: (a) x (t ) = α t + β t , vx (t ) = dx = 2α t + 3β t At t = 4.00 s: dt vx = 2(0.200 m/s )(4.00 s) + 3(0.0200 m/s3 )(4.00 s) = 2.56 m/s (b) a x (t ) = dvx = 2α + β t , so Fx = max = m(2α + β t ) At t = 4.00 s: dt Fx = (4.00 kg)[2(0.200 m/s ) + 6(0.0200 m/s3 )(4.00 s)] = 3.52 N © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6-25 (c) IDENTIFY and SET UP: Use Eq (6.6) to calculate the work EXECUTE: Wtot = K − K1 At t1 = 0, v1 = so K1 = Wtot = WF K = 12 mv22 = 12 (4.00 kg)(2.56 m/s)2 = 13.1 J Then Wtot = K − K1 gives that WF = 13.1 J EVALUATE: Since v increases with t, the kinetic energy increases and the work done is positive We can also calculate WF directly from Eq (6.7), by writing dx as vx dt and performing the integral 6.78 IDENTIFY and SET UP: Use Eq (6.6) You positive work and gravity does negative work Let point be at the base of the bridge and point be at the top of the bridge EXECUTE: (a) Wtot = K − K1 K1 = 12 mv12 = 12 (80.0 kg)(5.00 m/s)2 = 1000 J K = 12 mv22 = 12 (80.0 kg)(1.50 m/s) = 90 J Wtot = 90 J − 1000 J = −910 J (b) Neglecting friction, work is done by you (with the force you apply to the pedals) and by gravity: Wtot = Wyou + Wgravity The gravity force is w = mg = (80.0 kg)(9.80 m/s ) = 784 N, downward The displacement is 5.20 m, upward Thus φ = 180° and Wgravity = ( F cos φ ) s = (784 N)(5.20 m)cos180° = −4077 J Then Wtot = Wyou + Wgravity gives Wyou = Wtot − Wgravity = −910 J − (−4077 J) = +3170 J 6.79 EVALUATE: The total work done is negative and you lose kinetic energy IDENTIFY: The negative work done by the spring equals the change in kinetic energy of the car SET UP: The work done by a spring when it is compressed a distance x from equilibrium is − 12 kx K = EXECUTE: − 12 kx = K − K1 gives kx = 12 mv12 and k = (mv12 )/x = [(1200 kg)(0.65 m/s)2 ]/(0.090 m) = 6.3 × 104 N/m 6.80 EVALUATE: When the spring is compressed, the spring force is directed opposite to the displacement of the object and the work done by the spring is negative IDENTIFY: Apply Wtot = K − K1 SET UP: Let x0 be the initial distance the spring is compressed The work done by the spring is kx 2 − 12 kx , where x is the final distance the spring is compressed EXECUTE: (a) Equating the work done by the spring to the gain in kinetic energy, v= kx 2 = 12 mv , so 400 N/m k x0 = (0.060 m) = 6.93 m/s m 0.0300 kg (b) Wtot must now include friction, so mv 2 = Wtot = 12 kx02 − fx0 , where f is the magnitude of the friction force Then, v= k 2f 400 N/m 2(6.00 N) x0 − x0 = (0.060 m) − (0.060 m) = 4.90 m/s m m 0.0300 kg (0.0300 kg) (c) The greatest speed occurs when the acceleration (and the net force) are zero Let x be the amount the f 6.00 N = 0.0150 m spring is still compressed, so the distance the ball has moved is x0 − x kx = f , x = = k 400 N/m The ball is 0.0150 m from the end of the barrel, or 0.0450 m from its initial position To find the speed, the net work is Wtot = 12 k ( x02 − x ) − f ( x0 − x), so the maximum speed is vmax = k 2f ( x0 − x ) − ( x0 − x ) m m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-26 Chapter 400 N/m 2(6.00 N) ((0.060 m) − (0.0150 m) ) − (0.060 m − 0.0150 m) = 5.20 m/s (0.0300 kg) (0.0300 kg) vmax = 6.81 EVALUATE: The maximum speed with friction present (part (c)) is larger than the result of part (b) but smaller than the result of part (a) IDENTIFY and SET UP: Use Eq (6.6) Work is done by the spring and by gravity Let point be where the textbook is released and point be where it stops sliding x2 = since at point the spring is neither stretched nor compressed The situation is sketched in Figure 6.81 EXECUTE: Wtot = K − K1 K1 = , K = Wtot = Wfric + Wspr Figure 6.81 Wspr = 12 kx12 , where x1 = 0.250 m (Spring force is in direction of motion of block so it does positive work.) Wfric = − μk mgd Then Wtot = K − K1 gives d= 6.82 kx12 − μk mgd = (250 N/m) (0.250 m) = 1.1 m, measured from the point where the block was released 2(0.30) (2.50 kg) (9.80 m/s ) EVALUATE: The positive work done by the spring equals the magnitude of the negative work done by friction The total work done during the motion between points and is zero and the textbook starts and ends with zero kinetic energy IDENTIFY: Apply Wtot = K − K1 to the cat μk mg = kx 2 SET UP: Let point be at the bottom of the ramp and point be at the top of the ramp EXECUTE: The work done by gravity is Wg = − mgL sin θ (negative since the cat is moving up), and the work done by the applied force is FL, where F is the magnitude of the applied force The total work is Wtot = (100 N)(2.00 m) − (7.00 kg)(9.80 m/s )(2.00 m)sin 30° = 131.4 J The cat’s initial kinetic energy is v2 = mv 2 = 12 (7.00 kg) (2.40 m/s)2 = 20.2 J, and 2( K1 + W ) 2(20.2 J + 131.4 J) = = 6.58 m/s m (7.00 kg) EVALUATE: The net work done on the cat is positive and the cat gains speed Without your push, Wtot = Wgrav = −68.6 J and the cat wouldn’t have enough initial kinetic energy to reach the top of the ramp 6.83 IDENTIFY: Apply Wtot = K − K1 to the vehicle SET UP: Call the bumper compression x and the initial speed v0 The work done by the spring is − 12 kx and K = EXECUTE: (a) The necessary relations are x, the two inequalities are x > x> (20.0 m/s) 5(9.80 m/s ) 2 kx = mv0 , kx < mg Combining to eliminate k and then 2 v2 mg and k < 25 Using the given numerical values, 5g v = 8.16 m and k < 25 (1700 kg) (9.80 m/s ) (20.0 m/s) = 1.02 × 104 N/m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6-27 (b) A distance of m is not commonly available as space in which to stop a car Also, the car stops only momentarily and then returns to its original speed when the spring returns to its original length EVALUATE: If k were doubled, to 2.04 × 104 N/m, then x = 5.77 m The stopping distance is reduced by a factor of 1/ 2, but the maximum acceleration would then be kx/m = 69.2 m/s , which is 7.07 g 6.84 IDENTIFY: Apply Wtot = K − K1 W = Fs cos φ SET UP: The students positive work, and the force that they exert makes an angle of 30.0° with the direction of motion Gravity does negative work, and is at an angle of 120.0° with the chair’s motion EXECUTE: The total work done is Wtot = ((600 N) cos30.0° + (85.0 kg)(9.80 m/s ) cos120.0°)(2.50 m) = 257.8 J, and so the speed at the top of the ramp is v2 = v12 + 2Wtot 2(257.8 J) = (2.00 m/s) + = 3.17 m/s m (85.0 kg) EVALUATE: The component of gravity down the incline is mg sin 30° = 417 N and the component of the push up the incline is (600 N)cos30° = 520 N The force component up the incline is greater than the force 6.85 component down the incline; the net work done is positive and the speed increases IDENTIFY: Apply Wtot = K − K1 to the blocks SET UP: If X is the distance the spring is compressed, the work done by the spring is − 12 kX At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy and x2 = EXECUTE: (a) The work done by the block is equal to its initial kinetic energy, and the maximum 5.00 kg m compression is found from 12 kX = 12 mv02 and X = (6.00 m/s) = 0.600 m v0 = 500 N/m k (b) Solving for v0 in terms of a known X, v0 = 6.86 k 500 N/m X= (0.150 m) = 1.50 m/s m 5.00 kg EVALUATE: The negative work done by the spring removes the kinetic energy of the block IDENTIFY: Apply Wtot = K − K1 to the system of the two blocks The total work done is the sum of that done by gravity (on the hanging block) and that done by friction (on the block on the table) SET UP: Let h be the distance the 6.00 kg block descends The work done by gravity is (6.00 kg)gh and the work done by friction is − μk (8.00 kg) gh EXECUTE: Wtot = (6.00 kg − (0.25)(8.00 kg))(9.80 m/s )(1.50 m) = 58.8 J This work increases the 2(58.8 J) kinetic energy of both blocks: Wtot = (m1 + m2 )v , so v = = 2.90 m/s (14.00 kg) 6.87 EVALUATE: Since the two blocks are connected by the rope, they move the same distance h and have the same speed v IDENTIFY and SET UP: Apply Wtot = K − K1 to the system consisting of both blocks Since they are connected by the cord, both blocks have the same speed at every point in the motion Also, when the 6.00-kg block has moved downward 1.50 m, the 8.00-kg block has moved 1.50 m to the right The target variable, μk , will be a factor in the work done by friction The forces on each block are shown in Figure 6.87 EXECUTE: K1 = 12 m Av12 + 12 mB v12 = 12 (m A + mB )v12 K2 = Figure 6.87 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-28 Chapter The tension T in the rope does positive work on block B and the same magnitude of negative work on block A, so T does no net work on the system Gravity does work Wmg = m A gd on block A, where d = 2.00 m (Block B moves horizontally, so no work is done on it by gravity.) Friction does work Wfric = − μk mB gd on block B Thus Wtot = Wmg + Wfric = m A gd − μk mB gd Then Wtot = K − K1 gives m A gd − μk mB gd = − 12 (m A + mB )v12 and μk = m A 12 (m A + mB )v1 6.00 kg (6.00 kg + 8.00 kg) (0.900 m/s) + = + = 0.786 mB mB gd 8.00 kg 2(8.00 kg) (9.80 m/s ) (2.00 m) EVALUATE: The weight of block A does positive work and the friction force on block B does negative work, so the net work is positive and the kinetic energy of the blocks increases as block A descends Note that K1 includes the kinetic energy of both blocks We could have applied the work-energy theorem to block A alone, but then Wtot includes the work done on block A by the tension force 6.88 IDENTIFY: Apply Wtot = K − K1 The work done by the force from the bow is the area under the graph of Fx versus the draw length SET UP: One possible way of estimating the work is to approximate the F versus x curve as a parabola which goes to zero at x = and x = x0 , and has a maximum of F0 at x = x0 /2, so that F ( x) = (4 F0 /x02 ) x ( x0 − x) This may seem like a crude approximation to the figure, but it has the advantage of being easy to integrate EXECUTE: x0 ∫0 Fdx = F0 x02 x0 ∫0 ( x0 x − x ) dx = F0 ⎛ x02 x03 ⎞ − ⎟ = F0 x0 With F0 = 200 N and ⎜ x0 ⎟⎠ x02 ⎜⎝ x0 = 0.75 m, W = 100 J The speed of the arrow is then 6.89 2W 2(100 J) = = 89 m/s m (0.025 kg ) EVALUATE: We could alternatively represent the area as that of a rectangle 180 N by 0.55 m This gives W = 99 J, in close agreement with our more elaborate estimate IDENTIFY: Apply Eq (6.6) to the skater SET UP: Let point be just before she reaches the rough patch and let point be where she exits from the patch Work is done by friction We don’t know the skater’s mass so can’t calculate either friction or the initial kinetic energy Leave her mass m as a variable and expect that it will divide out of the final equation EXECUTE: f k = 0.25mg so W f = Wtot = −(0.25mg )s, where s is the length of the rough patch Wtot = K − K1 K1 = 12 mv02 , K = 12 mv22 = 12 m(0.55v0 ) = 0.3025 ( 12 mv02 ) The work-energy relation gives −(0.25mg ) s = (0.3025 − 1) 12 mv02 6.90 The mass divides out, and solving gives s = 1.3 m EVALUATE: Friction does negative work and this reduces her kinetic energy IDENTIFY: Pav = F||vav Use F = ma to calculate the force SET UP: vav = + 6.00 m/s = 3.00 m/s v − v0 6.00 m/s = = 2.00 m/s Since there are no t 3.00 s other horizontal forces acting, the force you exert on her is given by Fnet = ma = (65.0 kg)(2.00 m/s ) = 130 N Pav = (130 N)(3.00 m/s) = 390 W EXECUTE: Your friend’s average acceleration is a = EVALUATE: We could also use the work-energy theorem: W = K − K1 = 12 (65.0 kg)(6.00 m/s)2 = 1170 J Pav = W 1170 J = = 390 W, the same as obtained by our other approach t 3.00 s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6.91 6-29 IDENTIFY: To lift a mass m a height h requires work W = mgh To accelerate mass m from rest to speed v requires W = K − K1 = 12 mv Pav = SET UP: t = 60 s ΔW Δt EXECUTE: (a) (800 kg)(9.80 m/s )(14.0 m) = 1.10 × 105 J (b) (1/2)(800 kg)(18.0 m/s ) = 1.30 × 105 J 1.10 × 105 J + 1.30 × 105 J = 3.99 kW 60 s EVALUATE: Approximately the same amount of work is required to lift the water against gravity as to accelerate it to its final speed IDENTIFY and SET UP: W = Pt EXECUTE: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s At 10 beats/s, the bird (c) 6.92 must expend between 0.07 J/beat and 0.175 J/beat (b) The steady output of the athlete is (500 W)/(70 kg) = W/kg, which is below the 10 W/kg necessary to stay aloft Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human- 6.93 powered aircraft could stay aloft for very long EVALUATE: Movies of early attempts at human-powered flight bear out our results IDENTIFY and SET UP: Energy is Pavt The total energy expended in one day is the sum of the energy expended in each type of activity EXECUTE: day = 8.64 × 104 s Let t walk be the time she spends walking and tother be the time she spends in other activities; tother = 8.64 × 104 s − twalk The energy expended in each activity is the power output times the time, so E = Pt = (280 W)twalk + (100 W)tother = 1.1 × 107 J (280 W)twalk + (100 W)(8.64 × 104 s − t walk ) = 1.1 × 107 J (180 W)twalk = 2.36 × 106 J twalk = 1.31 × 104 s = 218 = 3.6 h EVALUATE: Her average power for one day is (1.1 × 107 J)/([ 24][3600 s]) = 127 W This is much closer to 6.94 her 100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate IDENTIFY and SET UP: Use Eq (6.15) The work done on the water by gravity is mgh, where h = 170 m Solve for the mass m of water for 1.00 s and then calculate the volume of water that has this mass ΔW EXECUTE: The power output is Pav = 2000 MW = 2.00 × 109 W Pav = and 92% of the work done Δt on the water by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is P Δt (2.00 × 109 W)(1.00 s ) W = av = = 2.174 × 109 J 0.92 0.92 W = mgh, so the mass of water flowing over the dam in 1.00 s must be m= W 2.174 × 109 J = = 1.30 × 106 kg gh (9.80 m/s )(170 m) density = m m 1.30 × 106 kg = = 1.30 × 103 m3 so V = density 1.00 × 103 kg/m3 V EVALUATE: The dam is 1270 m long, so this volume corresponds to about a m3 flowing over each m length of the dam, a reasonable amount © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-30 6.95 Chapter IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by the heart in one day For part (b) use W calculated in part (a) in Eq (6.15) EXECUTE: (a) W = mgh, as in Example 6.10 We need the mass of blood lifted; we are given the volume ⎛ × 10−3 m3 ⎞ V = (7500 L) ⎜ ⎟⎟ = 7.50 m ⎜ L ⎝ ⎠ m = density × volume = (1.05 × 103 kg/m3 )(7.50 m3 ) = 7.875 × 103 kg Then W = mgh = (7.875 × 103 kg)(9.80 m/s )(1.63 m) = 1.26 × 105 J ΔW 1.26 × 105 J = = 1.46 W Δt (24 h)(3600 s/h) EVALUATE: Compared to light bulbs or common electrical devices, the power output of the heart is rather small IDENTIFY: P = F&v = Mav To overcome gravity on a slope that is at an angle α above the horizontal, (b) Pav = 6.96 P = (Mg sin α )v SET UP: MW = 106 W kN = 103 N When α is small, tan α ≈ sin α EXECUTE: (a) The number of cars is the total power available divided by the power needed per car, 13.4 × 106 W = 177, rounding down to the nearest integer (2.8 × 103 N)(27 m/s) (b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav To climb a hill of angle α , the extra power needed is ( Mg sin α )v This will be nearly the same if a ~ g sin α ; if g sin α ~ g tan α ~ 0.10 m/s , the power is about the same as that needed to accelerate at 0.10 m/s (c) P = ( Mg sin α )v, where M is the total mass of the diesel units P = (1.10 × 106 kg)(9.80 m/s )(0.010)(27 m/s) = 2.9 MW (d) The power available to the cars is 13.4 MW, minus the 2.9 MW needed to maintain the speed of the diesel units on the incline The total number of cars is then 13.4 × 106 W − 2.9 × 106 W = 36, rounding to the nearest integer (2.8 × 103 N + (8.2 × 104 kg)(9.80 m/s )(0.010))(27 m/s) EVALUATE: For a single car, Mg sin α = (8.2 × 104 kg)(9.80 m/s )(0.010) = 8.0 × 103 N, which is over 6.97 twice the 2.8 kN required to pull the car at 27 m/s on level tracks Even a slope as gradual as 1.0% greatly increases the power requirements, or for constant power greatly decreases the number of cars that can be pulled IDENTIFY: P = F||v The force required to give mass m an acceleration a is F = ma For an incline at an angle α above the horizontal, the component of mg down the incline is mg sin α SET UP: For small α , sin α ≈ tan α EXECUTE: (a) P0 = Fv = (53 × 103 N)(45 m/s) = 2.4 MW (b) P1 = mav = (9.1 × 105 kg)(1.5 m/s )(45 m/s) = 61 MW (c) Approximating sinα , by tanα , and using the component of gravity down the incline as mgsinα , P2 = (mgsinα )v = (9.1 × 105 kg)(9.80 m/s )(0.015)(45 m/s) = 6.0 MW EVALUATE: From Problem 6.96, we would expect that a 0.15 m/s acceleration and a 1.5% slope would require the same power We found that a 1.5 m/s acceleration requires ten times more power than a 1.5% slope, which is consistent 6.98 x2 F dx, x1 x IDENTIFY: W = ∫ and Fx depends on both x and y SET UP: In each case, use the value of y that applies to the specified path ∫ xdx = 12 x ∫ x dx = 13 x 2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6-31 EXECUTE: (a) Along this path, y is constant, with the value y = 3.00 m (2.00 m) = 15.0 J, since x1 = and x2 = 2.00 m (b) Since the force has no y-component, no work is done moving in the y-direction (c) Along this path, y varies with position along the path, given by y = 1.5 x, so Fx = α (1.5 x) x = 1.5α x , and x2 x1 W = α y∫ xdx = (2.50 N/m )(3.00 m) (2.00 m)3 = 10.0 J EVALUATE: The force depends on the position of the object along its path IDENTIFY and SET UP: Use Eq (6.18) to relate the forces to the power required The air resistance force is Fair = 12 CAρ v , where C is the drag coefficient x2 Fdx x1 W =∫ 6.99 x2 x dx x1 = 1.5α ∫ = 1.5(2.50 N/m ) EXECUTE: (a) P = Ftot v, with Ftot = Froll + Fair Fair = 12 CAρ v = 12 (1.0)(0.463 m3 )(1.2 kg/m3 )(12.0 m/s) = 40.0 N Froll = μ r n = μr w = (0.0045)(490 N + 118 N) = 2.74 N P = ( Froll + Fair ) v = (2.74 N + 40.0 N)(12.0 s) = 513 W (b) Fair = 12 CAρ v = 12 (0.88)(0.366 m3 )(1.2 kg/m3 )(12.0 m/s) = 27.8 N Froll = μr n = μr w = (0.0030)(490 N + 88 N) = 1.73 N P = ( Froll + Fair )v = (1.73 N + 27.8 N)(12.0 s) = 354 W (c) Fair = 12 CAρ v = 12 (0.88)(0.366 m3 )(1.2 kg/m3 )(6.0 m/s) = 6.96 N Froll = μr n = 1.73 N (unchanged) P = ( Froll + Fair )v = (1.73 N + 6.96 N)(6.0 m/s) = 52.1 W EVALUATE: Since Fair is proportional to v and P = Fv, reducing the speed greatly reduces the power 6.100 required IDENTIFY: P = F||v SET UP: m/s = 3.6 km/h P 28.0 × 103 W = = 1.68 × 103 N v (60.0 km/h)((1 m/s)/(3.6 km/h)) (b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35/4), and so the power at the lower speed is (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp (c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 + 0.35 × 4) = 114.8 kW = 154 hp EXECUTE: (a) F = 6.101 EVALUATE: At low speeds rolling friction dominates the power requirement but at high speeds air resistance dominates IDENTIFY and SET UP: Use Eq (6.18) to relate F and P In part (a), F is the retarding force In parts (b) and (c), F includes gravity EXECUTE: (a) P = Fv, so F = P/v ⎛ 746 W ⎞ P = (8.00 hp) ⎜ ⎟ = 5968 W ⎝ hp ⎠ ⎛ 1000 m ⎞⎛ h ⎞ v = (60.0 km/h) ⎜ ⎟⎜ ⎟ = 16.67 m/s ⎝ km ⎠⎝ 3600 s ⎠ P 5968 W F= = = 358 N v 16.67 m/s (b) The power required is the 8.00 hp of part (a) plus the power Pg required to lift the car against gravity The situation is sketched in Figure 6.101 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 6-32 Chapter 10 m = 0.10 100 m α = 5.71° tan α = Figure 6.101 The vertical component of the velocity of the car is v sin α = (16.67 m/s) sin 5.71° = 1.658 m/s Then Pg = F (v sin a ) = mgv sin α = (1800 kg)(9.80 m/s )(1.658 m/s) = 2.92 × 104 W ⎛ hp ⎞ Pg = 2.92 × 104 W ⎜ ⎟ = 39.1 hp ⎝ 746 W ⎠ The total power required is 8.00 hp + 39.1 hp = 47.1 hp (c) The power required from the engine is reduced by the rate at which gravity does positive work The road incline angle α is given by tan α = 0.0100, so α = 0.5729° Pg = mg (v sin α ) = (1800 kg) (9.80 m/s )(16.67 m/s) sin 0.5729° = 2.94 × 103 W = 3.94 hp The power required from the engine is then 8.00 hp − 3.94 hp = 4.06 hp (d) No power is needed from the engine if gravity does work at the rate of Pg = 8.00 hp = 5968 W Pg = mgv sin α , so sin α = Pg 5968 W = 0.02030 (1800 kg)(9.80 m/s )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade EVALUATE: More power is required when the car goes uphill and less when it goes downhill In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the 6.102 mgv = retarding force and no force from the engine is required The retarding force depends on the speed so it is the same in parts (a), (b) and (c) IDENTIFY: Apply Wtot = K − K1 to relate the initial speed v0 to the distance x along the plank that the box moves before coming to rest SET UP: The component of weight down the incline is mg sin α , the normal force is mg cos α and the friction force is f = μ mg cos α x EXECUTE: ΔK = − mv02 and W = ∫ (− mg sinα − μ mg cosα ) dx Then, x ⎡ ⎤ Ax W = 2mg ∫ (sinα + Ax cosα )dx, W = − mg ⎢sinα x + cosα ⎥ ⎥⎦ ⎣⎢ ⎡ ⎤ Ax cosα ⎥ To eliminate x, note that the box comes to a rest when Set W = ΔK : − mv02 = −mg ⎢sinα x + 2 ⎢⎣ ⎥⎦ the force of static friction balances the component of the weight directed down the plane So, sinα mg sin α = Ax mg cos α Solve this for x and substitute into the previous equation: x = Then, Acosα ⎡ ⎤ sinα ⎛ sinα ⎞ ⎢ v0 = + g sinα + A⎜ ⎟ cosα ⎥ , and upon canceling factors and collecting terms, Acosα ⎝ Acosα ⎠ ⎢ ⎥ ⎣ ⎦ v02 = g sin α g sin α The box will remain stationary whenever v02 = Acosα Acosα EVALUATE: If v0 is too small the box stops at a point where the friction force is too small to hold the box in place sin α increases and cos α decreases as α increases, so the v0 required increases as α increases © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Work and Kinetic Energy 6.103 6-33 IDENTIFY: In part (a) follow the steps outlined in the problem For parts (b), (c) and (d) apply the workenergy theorem SET UP: ∫ x 2dx = 13 x3 EXECUTE: (a) Denote the position of a piece of the spring by l; l = is the fixed point and l = L is the moving end of the spring Then the velocity of the point corresponding to l, denoted u, is u (l ) = v(l / L) (when the spring is moving, l will be a function of time, and so u is an implicit function of time) The mass of a piece 1 Mv 2 Mv L Mv and , K = dK = l dl = of length dl is dm = ( M/L) dl , and so dK = (dm)u = l dl ∫ ∫ 2 L3 L3 (b) kx 2 = 12 mv , so v = (k/m) x = (3200 N/m)/(0.053 kg) (2.50 × 10−2 m) = 6.1 m/s (c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so 12 kx = 12 mv + 16 Mv Solving for v, v= (d) Algebraically, (3200 N/m) k (2.50 × 10−2 m) = 3.9 m/s x= (0.053 kg) + (0.243 kg)/3 m + M/3 (1/2)kx (1/2) kx mv = = 0.40 J and Mv = = 0.60 J (1 + M /3m) (1 + 3m/M ) ⎛ 0.053 kg ⎞ 3m K ball = = 3⎜ ⎟ = 0.65 The percentage of the final Kspring M ⎝ 0.243 kg ⎠ kinetic energy that ends up with each object depends on the ratio of the masses of the two objects As expected, when the mass of the spring is a small fraction of the mass of the ball, the fraction of the kinetic energy that ends up in the spring is small IDENTIFY: In both cases, a given amount of fuel represents a given amount of work W0 that the engine EVALUATE: For this ball and spring, 6.104 does in moving the plane forward against the resisting force Write W0 in terms of the range R and speed v and in terms of the time of flight T and v SET UP: In both cases assume v is constant, so W0 = RF and R = vT β ⎞ ⎛ EXECUTE: In terms of the range R and the constant speed v, W0 = RF = R ⎜ α v + ⎟ v ⎠ ⎝ β⎞ ⎛ In terms of the time of flight T,R = vt , so W0 = vTF = T ⎜ α v3 + ⎟ v⎠ ⎝ (a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, dW0 dR dF dF dR = to obtain F+R = For the maximum range, = 0, so = Performing the setting dv dv dv dv dv dF = 2α v − β /v3 = 0, which is solved for differentiation, dv 1/4 1/ ⎛ 3.5 × 105 N ⋅ m /s ⎞ =⎜ = 32.9 m/s = 118 km/h ⎜ 0.30 N ⋅ s /m ⎟⎟ ⎝ ⎠ d (b) Similarly, the maximum time is found by setting ( Fv) = 0; performing the differentiation, dv ⎛β ⎞ v=⎜ ⎟ ⎝α ⎠ 1/4 ⎛ β ⎞ 3α v − β /v = v = ⎜ ⎟ ⎝ 3α ⎠ 1/4 ⎛ 3.5 × 105 N ⋅ m /s ⎞ =⎜ 2 ⎟ ⎜ ⎟ ⎝ 3(0.30 N ⋅ s /m ) ⎠ = 25 m/s = 90 km/h EVALUATE: When v = ( β /α )1/ , Fair has its minimum value Fair = αβ For this v, R1 = (0.50) R2 = (0.43) W0 αβ W0 αβ and T1 = (0.50)α −1/4 β −3/4 When v = ( β /3α )1/ , Fair = 2.3 αβ For this v, and T2 = (0.57)α −1/ β −3/4 R1 > R2 and T2 > T1, as they should be © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher