14–1. SOLUTION Equation of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is . Applying Eq. 13–7, we have Principle of Work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force does negative work since it acts in the opposite direction to that of displacement. The normal reaction N , the vertical component of force F and the weight of the crate do not displace hence do no work. Applying Eq.14–7, we have Ans. v = 10.7 m s - L 25 m 15 m 36.55 ds = 1 2 (20) v 2 1 2 (20)(8 2) + L 25 m 15 m 100 cos 30° ds T 1 + a U 1 - 2 = T 2 F f = 0.25(146.2) = 36.55 N N = 146.2 N + c a F y = ma y ; N + 100 sin 30° - 20(9.81) = 20(0) F f = m k N = 0.25 N 30 ° F The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m / s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is m k = 0.25. Ans: v = 10.7 m > s © Pearson Education Limited 2017. 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. 378 14–2. The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of The coefficient of kinetic friction between the crate and the surface is m k = 0.2 6 m > s. SOLUTION Equations of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is . Applying Eq. 13–7, we have Principle of W ork and Energy: The horizontal components of force 800 N and 1000 N which act in the direction of displacement do positive work, whereas the friction force does negative work since it acts in the opposite direction to that of displacement. The normal reaction N , the vertical component of 800 N and 1000 N force and the weight of the crate do not displace, hence they do no work. Since the crate is originally at rest, . Applying Eq. 14–7, we have Ans. s = 1.35m 0 + 800 cos 30°( s ) + 1000 a 4 5 b s - 156.2 s = 1 2 (100) A 6 2 B T 1 + a U 1 - 2 = T 2 T 1 = 0 F f = 0.2(781) = 156.2 N N = 781 N + c © F y = ma y ; N + 1000 a 3 5 b - 800 sin 30° - 100(9.81) = 100(0) F f = m k N = 0.2 N 3 4 5 1000 N 30 800 N Ans: s = 1.35 m © Pearson Education Limited 2017. 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. 379 14–3. The 100-kg crate is subjected to the forces shown. If it is originally at rest, determine the distance it slides in order to attain a speed of v = 8 m > s. The coefficient of kinetic friction between the crate and the surface is m k = 0.2. Solution Work. Consider the force equilibrium along the y axis by referring to the FBD of the crate, Fig. a , + c Σ F y = 0 ; N + 500 sin 45 ° - 100(9.81) - 400 sin 30 ° = 0 N = 827.45 N Thus, the friction is F f = m k N = 0.2(827.45) = 165.49 N . Here, F 1 and F 2 do positive work whereas F f does negative work. W and N do no work U F 1 = 400 cos 30 ° s = 346.41 s U F 2 = 500 cos 45 ° s = 353.55 s U F f = - 165.49 s Principle of Work And Energy. Applying Eq. 14–7 , T 1 + Σ U 1 - 2 = T 2 0 + 346.41 s + 353.55 s + ( - 165.49 s) = 1 2 (100)(8 2 ) s = 5.987 m = 5.99 m Ans. 400 N 30 45 500 N Ans: s = 5.99 m © Pearson Education Limited 2017. 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. 380 *14–4. Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A it will reach a speed of 100 km >h when it comes to the bottom Also, what should be the minimum radius of curvature r for the track at B so that the passengers do not experience a normal force greater than 4 mg = (39.24 m ) N? Neglect the size of the car and passenger. Solution 100 km >h = 100(10 3 ) 3600 = 27.778 m >s T 1 + Σ U 1 - 2 = T 2 0 + m (9.81) h = 1 2 m (27.778) 2 h = 39.3 m Ans. + c Σ F n = ma n ; 39.24 m - mg = m a (27.778) 2 r b r = 26.2 m Ans. A h B r Ans: h = 39.3 m r = 26.2 m © Pearson Education Limited 2017. 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. B 381 F ( N ) F 80 0(10) 3 x 1 / 2 x ( m ) SOLUTION Principle of Work and Energy: The speed of the car just before it crashes into the barrier is . The maximum penetration occurs when the car is brought to a stop, i.e., . Referring to the free-body diagram of the car, Fig. a , W and N do no work; however, does negative work. Ans. T 1 + © U 1 - 2 = T 2 F b v 2 = 0 v 1 = 20 m > s max 2 3 1/2 0 max 1 20 000 (20 ) 800(10 ) 0 2 9.81 0.825 m x x dx x ⎡ ⎤ ⎛ ⎞ + − = ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ = W = 20 000 N L © Pearson Education Limited 2017. 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. x max = 0.825 m Ans: 14–5. For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = [800(10 3 ) x 1 >2 ] N, where x is in m, determine the car’s maximum penetration in the barrier. The car has a mass of 2 Mg and it is traveling with a speed of 20 m >s just before it hits the barrier. 382 14–6. The force of F = 50 N is applied to the cord when s = 2 m. If the 6-kg collar is orginally at rest, determine its velocity at s = 0. Neglect friction. Solution Work. Referring to the FBD of the collar, Fig. a , we notice that force F does positive work but W and N do no work. Here, the displacement of F is s = 2 2 2 + 1.5 2 - 1.5 = 1.00 m U F = 50(1.00) = 50.0 J Principle of Work And Energy. Applying Eq. 14–7 , T 1 + Σ U 1 - 2 = T 2 0 + 50 = 1 2 (6) v 2 v = 4.082 m > s = 4.08 m > s Ans. A s 1.5 m F Ans: v = 4.08 m > s © Pearson Education Limited 2017. 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. 383 14–7. A force of F = 250 N is applied to the end at B . Determine the speed of the 10-kg block when it has moved 1.5 m, starting from rest. Solution Work. with reference to the datum set in Fig. a , S W + 2 s F = l d S W + 2d s F = 0 (1) Assuming that the block moves upward 1.5 m, then d S W = - 1.5 m since it is directed in the negative sense of S W . Substituted this value into Eq. (1), - 1.5 + 2 d s F = 0 d s F = 0.75 m Thus, U F = F d S F = 250(0.75) = 187.5 J U W = - Wd S W = - 10(9.81)(1.5) = - 147.15 J Principle of Work And Energy. Applying Eq. 14–7 , T 1 + U 1 - 2 = T 2 0 + 187.5 + ( - 147.15) = 1 2 (10) v 2 v = 2.841 m > s = 2.84 m > s Ans. F A B Ans: v = 2.84 m > s © Pearson Education Limited 2017. 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. 384 *14–8. SOLUTION Observer A : Ans. Observer B : At , Block moves Thus Ans. Note that this result is less than that observed by A 2 m > s v 2 = 4.08 m > s 1 2(10)(3) 2 + 6(6.391) = 1 2(10) v 2 2 T 1 + © U 1 - 2 = T 2 10 - 3.609 = 6.391 m s ¿ = 2(1.805) = 3.609 m v = 2 m > s t = 1.805 s t 2 + 16.67 t - 33.33 = 0 10 = 0 + 5 t + 1 2(0.6) t 2 A : + B s = s 0 + v 0 t + 1 2 a c t 2 6 = 10 a a = 0.6 m > s 2 F = ma v 2 = 6.08 m > s 1 2(10)(5) 2 + 6(10) = 1 2(10) v 2 2 T 1 + © U 1 - 2 = T 2 As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x , determine the final speed of the block if it has an initial speed of and travels 10 m, both directed to the right and measured from the fi x ed frame. Compare the result with that obtained by an observer B , attached to the a x is and moving at a constant velocity of relative to A Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy. 2 m > s x ¿ 5 m > s 6 N 5 m / s 2 m / s 10 m B x x ¿ A Ans: Observer A : v 2 = 6.08 m > s Observer B : v 2 = 4.08 m > s © Pearson Education Limited 2017. 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. 385 14–9. When the driver applies the brakes of a light truck traveling 40 km >h, it skids 3 m before stopping. How far will the truck skid if it is traveling 80 km >h when the brakes are applied? Solution 40 km >h = 40(10 3 ) 3600 = 11.11 m >s 80 km >h = 22.22 m >s T 1 + Σ U 1 - 2 = T 2 1 2 m (11.11) 2 - m k mg (3) = 0 m k g = 20.576 T 1 + Σ U 1 - 2 = T 2 1 2 m (22.22) 2 - (20.576) m ( d ) = 0 d = 12 m Ans. Ans: d = 12 m © Pearson Education Limited 2017. 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. 386 14–10. The “air spring” A is used to protect the support B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D . The force developed by the air spring as a function of its deflection is shown by the graph. If the block has a mass of 20 kg and is suspended a height d = 0.4 m above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt. Solution Work. Referring to the FBD of the tensioning weight, Fig. a , W does positive work whereas force F does negative work. Here the weight displaces downward S W = 0.4 + x max where x max is the maximum compression of the air spring. Thus U W = 20(9.81)(0.4 + x max ) = 196.2(0.4 + x max ) The work of F is equal to the area under the F - S graph shown shaded in Fig. b , Here F x max = 1500 0.2 ; F = 7500 x max . Thus U F = - 1 2 (7500 x max )( x max ) = - 3750 x 2 max Principle of Work And Energy. Since the block is at rest initially and is required to stop momentarily when the spring is compressed to the maximum, T 1 = T 2 = 0. Applying Eq. 14–7 , T 1 + Σ U 1 - 2 = T 2 0 + 196.2(0.4 + x max ) + ( - 3750 x 2 max ) = 0 3750 x 2 max - 196.2x max - 78.48 = 0 x max = 0.1732 m = 0.173 m 6 0.2 m (O.K!) Ans. d B A D F (N) s (m) C 1500 0.2 Ans: x max = 0.173 m © Pearson Education Limited 2017. 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. 387 14–11. The force F , acting in a constant direction on the 20-kg block, has a magnitude which varies with the position s of the block. Determine how far the block must slide before its velocity becomes 15 m >s. When s = 0 the block is moving to the right at v = 6 m > s. The coefficient of kinetic friction between the block and surface is m k = 0.3. Solution Work. Consider the force equilibrium along y axis, by referring to the FBD of the block, Fig. a , + c Σ F y = 0 ; N - 20(9.81) = 0 N = 196.2 N Thus, the friction is F f = m k N = 0.3(196.2) = 58.86 N . Here, force F does positive work whereas friction F f does negative work. The weight W and normal reaction N do no work. U F = L Fds = L s 0 50 s 1 2 ds = 100 3 s 3 2 U F f = - 58.86 s Principle of Work And Energy. Applying Eq. 14–7 , T 1 + Σ U 1 - 2 = T 2 1 2(20)(6 2 ) + 100 3 s 3 2 + ( - 58.86 s ) = 1 2 (20)(15 2 ) 100 3 s 3 2 - 58.86 s - 1890 = 0 Solving numerically, s = 20.52 m = 20.5 m Ans. F (N) F 50 s 1 / 2 s (m) F v Ans: s = 20.5 m © Pearson Education Limited 2017. 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. 388 *14–12. SOLUTION Ans. Eliminating t , Solving for the positive root Ans. s = 130 m s 2 - 122.67 s - 981.33 = 0 s sin 30° + 4 = 0 + 0 + 1 2 (9.81) t 2 A + T B s = s 0 + v 0 t + 1 2 a c t 2 s cos 30° = 0 + 30.04 t A : + B s = s 0 + v 0 t v B = 30.04 m > s = 30.0 m > s 0 + 70(9.81)(46) = 1 2 (70)( v B ) 2 T A + © U A - B = T B The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed when he reaches B . Also, find the distance s to where he strikes the ground at C , if he makes the jump traveling horizontally at B . Neglect the skier’s size. He has a mass of 70 kg. v B 30 s 50 m 4 m A B C © Pearson Education Limited 2017. 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. s = 130 m v B = 30.0 m > s Ans: 389 14–13. SOLUTION Ans. k = 15.0 MN > m 2 40 000 - k (0.2) 3 3 = 0 1 2(5000)(4) 2 — L 0.2 0 ks 2 ds = 0 Design considerations for the bumper B on the 5-Mg train car require use of a nonlinear spring having the load- deflection characteristics shown in the graph. Select the proper value of k so that the ma x imum deflection of the spring is limited to 0.2 m when the car, traveling at strikes the rigid stop. Neglect the mass of the car wheels. 4 m > s, F (N) F ks 2 s (m) B Ans: k = 15.0 MN >m 2 © Pearson Education Limited 2017. 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. 390 A B 14–14. The 8-kg cylinder A and 3-kg cylinder B are released from rest. Determine the speed of A after it has moved 2 m starting from rest. Neglect the mass of the cord and pulleys. SOLUTION Kinematics: Express the length of cord in terms of position coordinates and by referring to Fig. a (1) Thus (2) If we assume that cylinder A is moving downward through a distance of , Eq. (2) gives Taking the time derivative of Eq. (1), (3) Positive net work on left means assumption of A moving down is correct. Since , Ans. v B = - 3.96 m > s = 3.96 m > s c v A = 1.98 m > s T v B = - 2 v A 0 + 8(2)9.81 - 3(4)9.81 = 1 2(8) v A 2 + 1 2(3) v B 2 g T 1 + g U 1 - 2 = g T 2 2 v A + v B = 0 ( + T ) ¢ s B = - 4 m = 4 m c 2(2) + ¢ s B = 0 ( + T ) ¢ s A = 2 m 2 ¢ s A + ¢ s B = 0 2 s A + s B = l s B s A © Pearson Education Limited 2017. 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. v A = 1.98 m > s T v B = 3.96 m > s c Ans: 391 SOLUTION Also, , and because the pulleys are massless, . The and terms drop out and the work-energy equation reduces to Ans. v A = 3.82 m > s 255.06 = 17.5 v 2 A F 2 F 1 F 1 = 2 F 2 v B = 2 v A 0 + 2[ F 1 - 3(9.81)] + 4[8(9.81) - F 2] = 1 2 (3) v 2 A + 1 2(8) v 2 B a T 1 + a U 1 - 2 = a T 2 A B © Pearson Education Limited 2017. 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. v A = 3.82 m > s Ans: 14–15. Cylinder A has a mass of 3 kg and cylinder B has a mass of 8 kg. Determine the speed of A after it has moved 2 m starting from rest. Neglect the mass of the cord and pulleys. 392 *14–16. SOLUTION Ans. v C = 2.36 m > s - 1 2 (25)(0.5 - 0.3) 2 = 1 2(20) v 2 C 0 + 100 sin 60°(0.5 - 0.3) + 196.2(0.5 - 0.3) - 1 2 (15)(0.5 - 0.3) 2 T 1 + a U 1 - 2 = T 2 The collar has a mass of 20 kg and is supported on the smooth rod. The attached springs are undeformed when Determine the speed of the collar after the applied force causes it to be displaced so that When the collar is at rest. d = 0.5 m d = 0.3 m. F = 100 N d = 0.5 m. 60 ° d k = 25 N/m k' = 15 N/m F = 100 N © Pearson Education Limited 2017. 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. v C = 2.36 m > s Ans: 393 14–17. SOLUTION Principle of Work and Energy: By referring to the free-body diagram of the block, Fig. a , notice that N does no work, while W does positive work since it displaces downward though a distance of (1) Equations of Motion: Here, . By referring to Fig. a , It is required that the block leave the track. Thus, Since , Ans. u = 41.41° = 41.4° 3 cos u - 9 4 = 0 mg Z 0 0 = mg a 3 cos u - 9 4 b N = 0 N = mg a 3 cos u - 9 4 b © F n = ma n ; mg cos u - N = m c g a 9 4 - 2 cos u b d a n = v 2 r = gr a 9 4 - 2 cos u b r = g a 9 4 - 2 cos u b v 2 = gr a 9 4 - 2 cos u b 1 2 m a 1 4 gr b + mg ( r - r cos u ) = 1 2 m v 2 T 1 + © U 1 - 2 = T 2 h = r - r cos u A small box of mass m is given a speed of at the top of the smooth half cylinder. Determine the angle at which the box leaves the cylinder. u v = 2 1 4 gr r O A u Ans: u = 41.4 ° © Pearson Education Limited 2017. 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. 394 14–18. A B C 300 mm 200 mm 200 mm 200 mm F 300 N 30 If the cord i s s ubjected to a con s tant force of and the 15-k g s mooth collar s tart s from re s t at A , determine the v elocity of the collar when it reache s point B . Ne g lect the s ize of the pulley. F = 300 N SOLUTION Free-Body D i a g ram : The free-body dia g ram of the collar and cord s y s tem at an arbitrary po s ition i s s hown in Fi g a Pr i nc ip le of Work and Ener g y : Referrin g to Fi g a , only N doe s no work s ince it alway s act s perpendicular to the motion. When the collar mo v e s from po s ition A to po s ition B , W di s place s v ertically upward a di s tance , while force F di s place s a di s tance of . Here, the work of F i s po s iti v e, wherea s W doe s ne g ati v e work. A n s v B = 3.335 m > s = 3.34 m > s 0 + 300(0.5234) + [ - 15(9.81)(0.5)] = 1 2 (15) v B 2 T A + g U A - B = T B 2 0.22 + 0.2 2 = 0.5234 m s = AC - BC = 2 0.7 2 + 0.4 2 - h = (0.3 + 0.2) m = 0.5 m Ans: v B = 3.34 m > s © Pearson Education Limited 2017. 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. 395 s M 14–19. If the force exerted by the motor M on the cable is 250 N, determine the speed of the 100-kg crate when it is hoisted to . The crate is at rest when s = 0 s = 3 m SOLUTION Kinematics: Expressing the length of the cable in terms of position coordinates and referring to Fig. a , (1) Using Eq. (1), the change in position of the crate and point P on the cable can be written as Here, . Thus, Principle of Work and Energy: Referring to the free-body diagram of the pulley system, Fig. b , and do no work since it acts at the support; however, T does positive work and does negative work. Ans. v = 1.07 m > s 0 + 250(12) + [ - 100(9.81)(3)] = 1 2 (100) v 2 0 + T ¢ s P + [ - W C ¢ s C ] = 1 2 m Cv 2 T 1 + g U 1 - 2 = T 2 W C F 2 F 1 ¢ s p = - 12 m = 12 m c 4( - 3) - ¢ s P = 0 ( + T ) ¢ s C = - 3 m 4 ¢ s C - ¢ s P = 0 ( + T ) 4 s C - s P = l 3 s C + ( s C - s P ) = l s P s C © Pearson Education Limited 2017. 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. v = 1.07 m > s Ans: 396 *14–20 . When a 7-kg projectile is fired from a cannon barrel that has a length of 2 m, the explosive force exerted on the projectile, while it is in the barrel, varies in the manner shown. Determine the approximate muzzle velocity of the projectile at the instant it leaves the barrel. Neglect the effects of friction inside the barrel and assume the barrel is horizontal. The work done is measured as the area under the force–displacement curve. This area is approximately 31.5 squares. Since each square has an area of , Ans. v 2 = 2121 m > s = 2.12 km > s (approx.) 0 + C (31.5)(2.5) A 106 B (0.2) D = 1 2 (7)( v 2 ) 2 T 1 + © U 1 - 2 = T 2 2.5 A 10 6 B (0.2) 15 10 5 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 F (MN) s (m) SOLUTION © Pearson Education Limited 2017. 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. v 2 = = 2.12 km > s Ans: 397