30 NEWTON'S LAWS [CHAP. 3 Solved Problems 3.1 Find the weight on Earth of a body whose mass is (a) 3.00 kg, (b) 200 g. The general relation between mass m and weight FW is FW mg. In this relation, m must be in kilograms, g in meters per second squared, and FW in newtons. On Earth, g 9:81 m/s2 . The acceleration due to gravity varies from place to place in the universe. 3.2 (a) FW 3:00 kg 9:81 m=s2 29:4 kgm=s2 29:4 N (b) FW 0:200 kg 9:81 m=s2 1:96 N A 20.0 kg object that can move freely is subjected to a resultant force of 45.0 N in the x-direction. Find the acceleration of the object. We make use of the second law in component form, Fx max , with Fx m 20:0 kg. Then Fx 45:0 N m 20:0 kg ax 2:25 N=kg 45:0 N and 2:25 m=s2 where we have used the fact that 1 N 1 kgm=s2 . Because the resultant force on the object is in the x-direction, its acceleration is also in that direction. 3.3 The object in Fig. 3-1(a) weighs 50 N and is supported by a cord. Find the tension in the cord. We mentally isolate the object for discussion. Two forces act on it, the upward pull of the cord and the downward pull of gravity. We represent the pull of the cord by FT , the tension in the cord. The pull of gravity, the weight of the object, is FW 50 N. These two forces are shown in the free-body diagram in Fig. 3-1(b). Fig. 3-1 The forces are already in component form and so we can write the ®rst condition for equilibrium at once, taking up and to the right as positive directions: ! Fx 0 becomes 00 " Fy 0 becomes FT 50 N 0 from which FT 50 N. Thus, when a single vertical cord supports a body at equilibrium, the tension in the cord equals the weight of the body. CHAP. 3] 3.4 31 NEWTON'S LAWS A 5.0 kg object is to be given an upward acceleration of 0.30 m/s2 by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. 3-2. The tension in the rope is FT , and the weight of the object is FW mg 5:0 kg 9:81 m=s2 49:1 N. Using Fy may with up taken as positive, we have FT mg may or FT 49:1 N 5:0 kg 0:30 m=s2 from which FT 50:6 N 51 N. As a check, we notice that FT is larger than FW as it must be if the object is to accelerate upward. Fig. 3-2 3.5 Fig. 3-3 A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal ¯oor at constant speed. What is the coecient of friction between ¯oor and box? Determine it to three signi®cant ®gures even though that's quite unrealistic. The free-body diagram for the box is shown in Fig. 3-3. Because the box does not move up or down, ay 0. Therefore, Fy may gives mg m 0 m=s2 FN from which we ®nd that FN mg 60:0 kg 9:81 m=s2 588:6 N. Further, because the box is moving horizontally at constant speed, ax 0 and so Fx max gives 140 N Ff 0 from which the friction force is Ff 140 N. We then have k 3.6 Ff 140 N 0:238 FN 588:6 N The only force acting on a 5.0 kg object has components Fx 20 N and Fy 30 N. Find the acceleration of the object. We make use of Fx max and Fy may to obtain 32 NEWTON'S LAWS [CHAP. 3 ax Fx 20 N 4:0 m=s2 m 5:0 kg ay Fy 30 N 6:0 m=s2 m 5:0 kg These components of the acceleration are shown in Fig. 3-4. From the ®gure, we see that q a 4:02 6:02 m=s2 7:2 m=s2 and arctan 6:0=4:0 568: Fig. 3-4 3.7 A 600 N object is to be given an acceleration of 0.70 m/s2 . How large an unbalanced force must act upon it? Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, we use FW mg to ®nd m FW 600 N 61 kg g 9:81 m=s2 Now that we know the mass of the object (61 kg) and the desired acceleration (0.70 m/s2 ), we have F ma 61 kg 0:70 m=s2 43 N 3.8 A constant force acts on a 5.0 kg object and reduces its velocity from 7.0 m/s to 3.0 m/s in a time of 3.0 s. Find the force. We must ®rst ®nd the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 we have a vf vi t 4:0 m=s 3:0 s 1:33 m=s2 Now we can use F ma with m 5:0 kg: F 5:0 kg 1:33 m=s2 6:7 N The minus sign indicates that the force is a retarding force, directed opposite to the motion. CHAP. 3] 3.9 33 NEWTON'S LAWS A 400-g block with an initial speed of 80 cm/s slides along a horizontal tabletop against a friction force of 0.70 N. (a) How far will it slide before stopping? (b) What is the coecient of friction between the block and the tabletop? (a) We take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, 0:70 N. Therefore, F ma becomes 0:70 N 0:400 kg a 2 from which a 1:75 m=s . (Notice that m is always in kilograms.) To ®nd the distance the block slides, we have vix 0:80 m/s, vfx 0, and a 1:75 m/s2 . Then v2fx v2ix 2ax gives x (b) v2ix 2a 0 0:64 m2 =s2 0:18 m 2 1:75 m=s2 Because the vertical forces on the block must cancel, the upward push of the table FN must equal the weight mg of the block. Then k 3.10 v2fx friction force 0:70 N 0:18 FN 0:40 kg 9:81 m=s2 A 600-kg car is moving on a level road at 30 m/s. (a) How large a retarding force (assumed constant) is required to stop it in a distance of 70 m? (b) What is the minimum coecient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction ± there's no sliding. (a) We must ®rst ®nd the car's acceleration from a motion equation. It is known that vix 30 m/s, vfx 0, and x 70 m. We use v2fx v2ix 2ax to ®nd a v2fx v2ix 2x 0 900 m2 =s2 140 m 6:43 m=s2 Now we can write F ma 600 kg 6:43 m=s2 (b) 3860 N 3:9 kN The force found in (a) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is Ff 3860 N. The coecient of friction is given by s Ff =FN , where FN is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, FN FW mg 600 kg 9:81 m=s2 5886 N F 3860 0:66 s f FN 5886 so that The coecient of friction must be at least 0.66 if the car is to stop within 70 m. 3.11 An 8000-kg engine pulls a 40 000-kg train along a level track and gives it an acceleration a1 1:20 m/s2 . What acceleration a2 would the engine give to a 16 000-kg train? For a given engine force, the acceleration is inversely proportional to the total mass. Thus a2 3.12 m1 8000 kg 40 000 kg a 1:20 m=s2 2:40 m=s2 m2 1 8000 kg 16 000 kg As shown in Fig. 3-5(a), an object of mass m is supported by a cord. Find the tension in the cord if the object is (a) at rest, (b) moving at constant velocity, (c) accelerating upward with acceleration a 3g=2, and (d ) accelerating downward at a 0:75g: 34 NEWTON'S LAWS [CHAP. 3 Two forces act on the object: the tension FT upward and the downward pull of gravity mg. They are shown in the free-body diagram in Fig. 3-5(b). We take up as the positive direction and write Fy may in each case. (a) ay 0: FT mg may 0 or FT mg (b) ay 0: FT mg may 0 or FT mg (c) ay 3g=2: FT mg m 3g=2 or FT 2:5mg (d ) ay FT mg m 3g=4 or FT 0:25mg 3g=4: Notice that the tension in the cord is less than mg in part (d ); only then can the object have a downward acceleration. Can you explain why FT 0 if ay g? Fig. 3-5 3.13 Fig. 3-6 A tow rope will break if the tension in it exceeds 1500 N. It is used to tow a 700-kg car along level ground. What is the largest acceleration the rope can give to the car? (Remember that 1500 has four signi®cant ®gures; see Appendix A.) The forces acting on the car are shown in Fig. 3-6. Only the x-directed force is of importance, because the y-directed forces balance each other. Indicating the positive direction with a sign and a little arrow we write, F max becomes 1500 N 700 kg a ! x from which a 2:14 m=s2 : 3.14 Compute the least acceleration with which a 45-kg woman can slide down a rope if the rope can withstand a tension of only 300 N. The weight of the woman is mg 45 kg 9:81 m=s2 441 N. Because the rope can support only 300 N, the unbalanced downward force F on the woman must be at least 441 N 300 N 141 N. Her minimum downward acceleration is then a 3.15 F 141 N 3:1 m=s2 m 45 kg A 70-kg box is slid along the ¯oor by a 400-N force as shown in Fig. 3-7. The coecient of friction between the box and the ¯oor is 0.50 when the box is sliding. Find the acceleration of the box. CHAP. 3] 35 NEWTON'S LAWS Fig. 3-7 Since the y-directed forces must balance, FN mg 70 kg 9:81 m=s2 687 N But the friction force Ff is given by Ff k FN 0:50 687 N 344 N Now write Fx max for the box, taking the direction of motion as positive: 400 N 3.16 344 N 70 kg a or a 0:80 m=s2 Suppose, as shown in Fig. 3-8, that a 70-kg box is pulled by a 400-N force at an angle of 308 to the horizontal. The coecient of kinetic friction is 0.50. Find the acceleration of the box. Fig. 3-8 Because the box does not move up or down, we have Fy may 0. From Fig. 3-8, we see that this equation is FN 200 N mg 0 But mg 70 kg 9:81 m=s2 687 N, and it follows that FN 486 N: We next ®nd the friction force acting on the box: Ff k FN 0:50 486 N 243 N Now let us write Fx max for the box. It is 346 from which ax 1:5 m/s2 : 243 N 70 kg ax 36 3.17 NEWTON'S LAWS [CHAP. 3 A car moving at 20 m/s along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coecient between tires and road is 0.90? Assume that all four wheels brake identically. If the brakes don't lock the car stops via static friction. The friction force at one wheel, call it wheel 1, is Ff1 s FN1 FW1 where FW1 is the weight carried by wheel 1. We obtain the total friction force Ff by adding such terms for all four wheels: Ff s FW1 s FW2 s FW3 s FW4 s FW1 FW2 FW3 FW4 s FW where FW is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This friction force is the only unbalanced force on the car (we neglect wind friction and such). Writing F ma for the car with F replaced by s FW gives s FW ma, where m is the car's mass and the positive direction is taken as the direction of motion. However, FW mg; so the car's acceleration is s F W s mg s g 0:90 9:81 m=s2 8:8 m=s2 m m We can ®nd how far the car went before stopping by solving a motion problem. Knowing that vi 20 m/s, vf 0, and a 8:8 m/s2 , we ®nd from v2f v2i 2ax that a x 400 m2 =s2 23 m 17:6 m=s2 0 If the four wheels had not all been braking optimally, the stopping distance would have been longer. 3.18 As shown in Fig. 3-9, a force of 400 N pushes on a 25-kg box. Starting from rest, the box achieves a velocity of 2.0 m/s in a time of 4.0 s. Find the coecient of kinetic friction between box and ¯oor. Fig. 3-9 We will need to ®nd f by use of F ma. But ®rst we must ®nd a from a motion problem. We know that vi 0, vf 2:0 m/s, t 4:0 s. Using vf vi at gives a vf vi t 2:0 m=s 0:50 m=s2 4:0 s Now we can write Fx max , where ax a 0:50 m/s2 . From Fig. 3-9, this equation becomes 257 N Ff 25 kg 0:50 m=s2 or Ff 245 N We now wish to use Ff =FN . To ®nd FN we write Fy may 0, since no vertical motion occurs. From Fig. 3-9, FN 306 N 25 9:81 N 0 or FN 551 N CHAP. 3] 37 NEWTON'S LAWS Then k 3.19 Ff 245 0:44 FN 551 A 200-N wagon is to be pulled up a 308 incline at constant speed. How large a force parallel to the incline is needed if friction eects are negligible? The situation is shown in Fig. 3-10(a). Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore the wagon is in translational equilibrium, and the ®rst condition for equilibrium applies to it. We isolate the wagon as the object. Three nonnegligible forces act on it: (1) the pull of gravity FW (its weight), directed straight down; (2) the force F exerted on the wagon parallel to the incline to pull it up the incline; (3) the push FN of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. 3-10(b). For situations involving inclines, it is convenient to take the x-axis parallel to the incline and the y-axis perpendicular to it. After taking components along these axes, we can write the ®rst condition for equilibrium: ! ! Fx 0 Fy 0 becomes F becomes FN 0:50 FW 0 0:87 FW 0 Solving the ®rst equation and recalling that FW 200 N, we ®nd that F 0:50 FW . The required pulling force to two signi®cant ®gures is 0.10 kN. Fig. 3-10 3.20 A 20-kg box sits on an incline as shown in Fig. 3-11. The coecient of kinetic friction between box and incline is 0.30. Find the acceleration of the box down the incline. In solving inclined-plane problems, we take x- and y-axes as shown in the ®gure, parallel and perpendicular to the incline. We shall ®nd the acceleration by writing Fx max . But ®rst we must ®nd the friction force Ff . Using the fact that cos 308 0:866, Fy may 0 gives FN 0:87mg 0 from which FN 0:87 20 kg 9:81 m=s2 171 N. Now we can ®nd Ff from Ff k FN 0:30 171 N 51 N 38 NEWTON'S LAWS [CHAP. 3 Fig. 3-11 Writing Fx max , we have Ff from which ax 3.21 0:50mg max or 51 N 0:50 20 9:81 N 20 kg ax 2:35 m/s2 . The box accelerates down the incline at 2.4 m/s2 . When a force of 500 N pushes on a 25-kg box as shown in Fig. 3-12, the acceleration of the box up the incline is 0.75 m/s2 . Find the coecient of kinetic friction between box and incline. The acting forces and their components are shown in Fig. 3-12. Notice how the x- and y-axes are taken. Since the box moves up the incline, the friction force (which always acts to retard the motion) is directed down the incline. Let us ®rst ®nd Ff by writing Fx max . From Fig. 3-12, using sin 408 0:643, 383 N Ff 0:64 25 9:81 N 25 kg 0:75 m=s2 from which Ff 207 N. We also need FN . Writing Fy may 0, and using cos 408 0:766, we get FN Then 321 N 0:77 25 9:81 N 0 k Ff 207 0:41 FN 510 Fig. 3-12 or FN 510 N CHAP. 3] 3.22 39 NEWTON'S LAWS Two blocks, of masses m1 and m2 , are pushed by a force F as shown in Fig. 3-13. The coecient of friction between each block and the table is 0.40. (a) What must be the value of F if the blocks are to have an acceleration of 200 cm/s2 ? How large a force does m1 then exert on m2 ? Use m1 300 g and m2 500 g. Remember to work in SI units. The friction forces on the blocks are Ff1 0:4m1 g and Ff2 0:4m2 g. We take the two blocks in combination as the object for discussion; the horizontal forces on the object from outside (i.e. the external forces on it) are F, Ff1 , and Ff2 . Although the two blocks do push on each other, the pushes are internal forces; they are not part of the unbalanced external force on the two-mass object. For that object, Fx max (a) becomes F Ff1 Ff2 m1 m2 ax Solving for F and substituting known values, we ®nd F 0:40 g m1 m2 m1 m2 ax 3:14 N 1:60 N 4:7 N (b) Now consider block m2 alone. The forces acting on it in the x-direction are the push of block m1 on it (which we represent by Fb ) and the retarding friction force Ff2 0:4m2 g. Then, for it, Fx max becomes Fb Ff2 m2 ax 2 We know that ax 2:0 m/s and so Fb Ff2 m2 ax 1:96 N 1:00 N 2:96 N 3:0 N Fig. 3-13 3.23 Fig. 3-14 A cord passing over an easily turned pulley (one that is both massless and frictionless) has a 7.0-kg mass hanging from one end and a 9.0-kg mass hanging from the other, as shown in Fig. 3-14. (This arrangement is called Atwood's machine.) Find the acceleration of the masses and the tension in the cord. Because the pulley is easily turned, the tension in the cord will be the same on each side. The forces acting on each of the two masses are drawn in Fig. 3-14. Recall that the weight of an object is mg. It is convenient in situations involving objects connected by cords to take the direction of motion as the positive direction. In the present case, we take up positive for the 7.0-kg mass, and down positive for the 9.0-kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn't stretch, the accelerations are numerically equal.) Writing Fy may for each mass in turn, we have FT 7:0 9:81 N 7:0 kg a and 9:0 9:81 N FT 9:0 kg a 40 NEWTON'S LAWS [CHAP. 3 If we add these two equations, the unknown FT drops out, giving 9:0 7:0 9:81 N 16 kg a 2 for which a 1:23 m/s . We can now substitute 1.23 m/s2 for a in either equation and obtain FT 77 N. 3.24 In Fig. 3-15, the coecient of kinetic friction between block A and the table is 0.20. Also, mA 25 kg, mB 15 kg. How far will block B drop in the ®rst 3.0 s after the system is released? Fig. 3-15 Since, for block A, there is no motion vertically, the normal force is FN mA g 25 kg 9:81 m=s2 245 N Ff k FN 0:20 245 N 49 N and We must ®rst ®nd the acceleration of the system and then we can describe its motion. Let us apply F ma to each block in turn. Taking the motion direction as positive, we have FT and mB g Ff mA a FT mB a or or FT 49 N 25 kg a FT 15 9:81 N 15 kg a We can eliminate FT by adding the two equations. Then, solving for a, we ®nd a 2:45 m/s2 : Now we can work a motion problem with a 2:45 m/s2 , vi 0, t 3:0 s: y viy t 12 at2 gives y 0 12 2:45 m=s2 3:0 s2 11 m as the distance B falls in the ®rst 3.0 s. 3.25 How large a horizontal force in addition to FT must pull on block A in Fig. 3-15 to give it an acceleration of 0.75 m/s2 toward the left? Assume, as in Problem 3.24, that k 0:20, mA 25 kg, and mB 15 kg. If we were to redraw Fig 3-15 for this case, we would show a force F pulling toward the left on A. In addition, the retarding friction force Ff should be reversed in direction in the ®gure. As in Problem 3.24, Ff 49 N. CHAP. 3] 41 NEWTON'S LAWS We write F ma for each block in turn, taking the direction of motion to be positive. We have F FT 49 N 25 kg 0:75 m=s2 and FT 15 9:81 N 15 kg 0:75 m=s2 We solve the last equation for FT and substitute in the previous equation. We can then solve for the single unknown F, and we ®nd it to be 226 N or 0.23 kN. 3.26 The coecient of static friction between a box and the ¯at bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one x-directed force, the friction force. When the box is on the verge of slipping, Ff s FW , where FW is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, Fx max applied to the box gives Ff max . However, if the box is on the verge of slipping, Ff s FW so that s FW max . Because FW mg, this gives mg s g 0:60 9:81 m=s2 5:9 m=s2 ax s m as the maximum acceleration without slipping. 3.27 In Fig. 3-16, the two boxes have identical masses of 40 kg. Both experience a sliding friction force with k 0:15. Find the acceleration of the boxes and the tension in the tie cord. Fig. 3-16 Using Ff FN , we ®nd that the friction forces on the two boxes are FfA 0:15 mg and FfB 0:15 0:87mg But m 40 kg, so FfA 59 N and FfB 51 N. Let us now apply Fx max to each block in turn, taking the direction of motion as positive. This gives FT 59 N 40 kg a and 0:5mg FT 51 N 40 kg a Solving these two equations for a and FT gives a 1:1 m=s2 and FT 0:10 kN. 42 3.28 NEWTON'S LAWS [CHAP. 3 In the system shown in Fig. 3-17(a), force F accelerates block m1 to the right. Find its acceleration in terms of F and the coecient of friction k at the contact surfaces. Fig. 3-17 The horizontal forces on the blocks are shown in Fig. 3-17(b) and (c). Block m2 is pressed against m1 by its weight m2 g. This is the normal force where m1 and m2 are in contact, so the friction force there is Ff 2 k m2 g. At the bottom surface of m1 , however, the normal force is m1 m2 g. Hence, Ff0 k m1 m2 g. We now write Fx max for each block, taking the direction of motion as positive: FT k m2 g m2 a and F FT m2 g k m1 m2 g m1 a We can eliminate FT by adding the two equations to obtain F 2k m2 g from which 3.29 k m1 m2 g m1 m2 a a F 2k m2 g m1 m2 k g In the system of Fig. 3-18, friction and the mass of the pulley are both negligible. Find the acceleration of m2 if m1 300 g, m2 500 g, and F 1:50 N. Fig. 3-18 Notice that m1 has twice as large an acceleration as m2 . (When the pulley moves a distance d, m1 moves a distance 2d.) Also notice that the tension FT1 in the cord pulling m1 is half FT 2 , that in the cord pulling the pulley, because the total force on the pulley must be zero. F ma tells us that this is so because the mass of the pulley is zero.) Writing Fx max for each mass, we have FT1 m1 2a and F FT2 m2 a However, we know that FT1 12 FT2 and so the ®rst equation gives FT2 4m1 a. Substitution in the second equation yields F 4m1 m2 a or a F 1:50 N 0:882 m=s2 4m1 m2 1:20 kg 0:50 kg CHAP. 3] 3.30 43 NEWTON'S LAWS In Fig. 3-19, the weights of the objects are 200 N and 300 N. The pulleys are essentially frictionless and massless. Pulley P1 has a stationary axle, but pulley P2 is free to move up and down. Find the tensions FT1 and FT2 and the acceleration of each body. Fig. 3-19 Mass B will rise and mass A will fall. You can see this by noticing that the forces acting on pulley P2 are 2FT2 up and FT1 down. Since the pulley has no mass, it can have no acceleration, and so FT1 2FT2 (the inertialess object transmits the tension). Twice as large a force is pulling upward on B as on A. Let a be the downward acceleration of A. Then a=2 is the upward acceleration of B. (Why?) We now write Fy may for each mass in turn, taking the direction of motion as positive in each case. We have FT1 300 N mB 12 a and 200 N FT2 mA a But m FW =g and so mA 200=9:81 kg and mB 300=9:81 kg. Further FT1 2FT2 . Substitution of these values in the two equations allows us to compute FT2 and then FT1 and a. The results are FT1 327 N 3.31 FT2 164 N a 1:78 m=s2 Compute the mass of the Earth, assuming it to be a sphere of radius 6370 km. Give your answer to three signi®cant ®gures. Let M be the mass of the Earth, and m the mass of an object on the Earth's surface. The weight of the object is equal to mg. It is also equal to the gravitational force G Mm=r2 , where r is the Earth's radius. Hence, mg G from which M Mm r2 gr2 9:81 m=s2 6:37 106 m2 5:97 1024 kg G 6:67 10 11 N m2 =kg2 44 NEWTON'S LAWS [CHAP. 3 Supplementary Problems 3.32 Once ignited, a small rocket motor on a spacecraft exerts a constant force of 10 N for 7.80 s. During the burn the rocket causes the 100-kg craft to accelerate uniformly. Determine that acceleration. Ans. 0.10 m/s2 3.33 Typically, a bullet leaves a standard 45-caliber pistol (5.0-in. barrel) at a speed of 262 m/s. If it takes 1 ms to traverse the barrel, determine the average acceleration experienced by the 16.2-g bullet within the gun and then compute the average force exerted on it. Ans. 3 105 m=s2 ; 0:4 102 N 3.34 A force acts on a 2-kg mass and gives it an acceleration of 3 m/s2 . What acceleration is produced by the same force when acting on a mass of (a) 1 kg? (b) 4 kg? (c) How large is the force? Ans. (a) 6 m/s2 ; (b) 2 m/s2 ; (c) 6 N 3.35 An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon? (c) What will be its acceleration on the Moon when a 0.500 N resultant force acts on it? Ans. (a) 2.94 N; (b) 0.300 kg; (c) 1.67 m/s2 3.36 A horizontal cable pulls a 200-kg cart along a horizontal track. The tension in the cable is 500 N. Starting from rest, (a) How long will it take the cart to reach a speed of 8.0 m/s? (b) How far will it have gone? Ans. (a) 3.2 s; (b) 13 m 3.37 A 900-kg car is going 20 m/s along a level road. How large a constant retarding force is required to stop it in a distance of 30 m? (Hint: First ®nd its deceleration.) Ans. 6.0 kN 3.38 A 12.0-g bullet is accelerated from rest to a speed of 700 m/s as it travels 20.0 cm in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? (Be careful of units.) Ans. 14.7 kN 3.39 A 20-kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is (a) 250 N, (b) 150 N, (c) zero, (d ) 196 N. Ans. (a) 2.7 m/s2 up; (b) 2.3 m/s2 down; 2 (c) 9.8 m/s down; (d ) zero 3.40 A 5.0-kg mass hangs at the end of a cord. Find the tension in the cord if the acceleration of the mass is Ans. (a) 57 N; (b) 42 N; (c) zero (a) 1.5 m/s2 up, (b) 1.5 m/s2 down, (c) 9.8 m/s2 down. 3.41 A 700-N man stands on a scale on the ¯oor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of (a) 1.8 m/s2 up? (b) 1.8 m/s2 down? (c) 9.8 m/s2 down? Ans. (a) 0.83 kN; (b) 0.57 kN; (c) zero 3.42 Using the scale described in Problem 3.41, a 65.0 kg astronaut weighs himself on the Moon, where Ans. 104 N g 1:60 m/s2 . What does the scale read? 3.43 A cord passing over a frictionless, massless pulley has a 4.0-kg object tied to one end and a 12-kg object tied to the other. Compute the acceleration and the tension in the cord. Ans. 4.9 m/s2 , 59 N 3.44 An elevator starts from rest with a constant upward acceleration. It moves 2.0 m in the ®rst 0.60 s. A passenger in the elevator is holding a 3.0-kg package by a vertical string. What is the tension in the string during the accelerating process? Ans. 63 N 3.45 Just as her parachute opens, a 60-kg parachutist is falling at a speed of 50 m/s. After 0.80 s has passed, the chute is fully open and her speed has dropped to 12.0 m/s. Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform. Ans. 2850 N 588 N 3438 N 3:4 kN CHAP. 3] 45 NEWTON'S LAWS 3.46 A 300-g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900-g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700 m/s2 : (b) Find the tension in each string when the acceleration is 0.700 m/s2 downward. Ans. (a) 12.6 N and 9.45 N; (b) 10.9 N and 8.19 N 3.47 A 20-kg wagon is pulled along the level ground by a rope inclined at 308 above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and (b) an acceleration of 0.40 m/s2 ? Ans. (a) 35 N; (b) 44 N 3.48 A 12-kg box is released from the top of an incline that is 5.0 m long and makes an angle of 408 to the horizontal. A 60-N friction force impedes the motion of the box. (a) What will be the acceleration of the box and (b) how long will it take to reach the bottom of the incline? Ans. (a) 1.3 m/s2 ; (b) 2.8 s 3.49 For the situation outlined in Problem 3.48, what is the coecient of friction between box and incline? Ans. 0.67 3.50 An inclined plane makes an angle of 308 with the horizontal. Find the constant force, applied parallel to the plane, required to cause a 15-kg box to slide (a) up the plane with acceleration 1.2 m/s2 and (b) down the incline with acceleration 1.2 m/s2 . Neglect friction forces. Ans. (a) 92 N; (b) 56 N 3.51 A horizontal force F is exerted on a 20-kg box to slide it up a 308 incline. The friction force retarding the motion is 80 N. How large must F be if the acceleration of the moving box is to be (a) zero and (b) 0.75 m/s2 ? Ans. (a) 0.21 kN; (b) 0.22 kN 3.52 An inclined plane making an angle of 258 with the horizontal has a pulley at its top. A 30-kg block on the plane is connected to a freely hanging 20-kg block by means of a cord passing over the pulley. Compute the distance the 20-kg block will fall in 2.0 s starting from rest. Neglect friction. Ans. 2.9 m 3.53 Repeat Problem 3.52 if the coecient of friction between block and plane is 0.20. 3.54 A horizontal force of 200 N is required to cause a 15-kg block to slide up a 208 incline with an acceleration of 25 cm/s2 . Find (a) the friction force on the block and (b) the coecient of friction. Ans. (a) 0.13 kN; (b) 0.65 3.55 Find the acceleration of the blocks in Fig. 3-20 if friction forces are negligible. What is the tension in the cord connecting them? Ans. 3.3 m/s2 , 13 N Ans. 0.74 m Fig. 3-20 3.56 Repeat Problem 3.55 if the coecient of kinetic friction between the blocks and the table is 0.30. Ans. 0.39 m/s2 , 13 N 3.57 How large a force F is needed in Fig. 3-21 to pull out the 6.0-kg block with an acceleration of 1.50 m/s2 if the coecient of friction at its surfaces is 0.40? Ans. 48 N 46 NEWTON'S LAWS Fig. 3-21 [CHAP. 3 Fig. 3-22 3.58 In Fig. 3-22, how large a force F is needed to give the blocks an acceleration of 3.0 m/s2 if the coecient of kinetic friction between blocks and table is 0.20? How large a force does the 1.50-kg block then exert on the 2.0-kg block? Ans. 22 N, 15 N 3.59 (a) What is the smallest force parallel to a 378 incline needed to keep a 100-N weight from sliding down the incline if the coecients of static and kinetic friction are both 0.30? (b) What parallel force is required to keep the weight moving up the incline at constant speed? (c) If the parallel pushing force is 94 N, what will be the acceleration of the object? (d ) If the object in (c) starts from rest, how far will it move in 10 s? Ans. (a) 36 N; (b) 84 N; (c) 0.98 m/s2 up the plane; (d ) 49 m 3.60 A 5.0-kg block rests on a 308 incline. The coecient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding (a) up the incline and (b) down the incline? Ans. (a) 43 N; (b) 16.6 N 3.61 Three blocks with masses 6.0 kg, 9.0 kg, and 10 kg are connected as shown in Fig. 3-23. The coecient of friction between the table and the 10-kg block is 0.20. Find (a) the acceleration of the system and (b) the tension in the cord on the left and in the cord on the right. Ans. (a) 0.39 m/s2 ; (b) 61 N, 85 N Fig. 3-23 3.62 The Earth's radius is about 6370 km. An object that has a mass of 20 kg is taken to a height of 160 km above the Earth's surface. (a) What is the object's mass at this height? (b) How much does the object weigh (i.e., how large a gravitational force does it experience) at this height? Ans. (a) 20 kg; (b) 0.19 kN 3.63 The radius of the Earth is about 6370 km, while that of Mars is about 3440 km. If an object weighs 200 N on Earth, what would it weigh, and what would be the acceleration due to gravity, on Mars? The mass of Mars is 0.11 that of Earth. Ans. 75 N, 3.7 m/s2