5 points

You can play catch with a softball in a bus moving with constant speed on a straight road, just as though the bus were at rest. Is this still possible when the bus is making a turn at constant speed on a level road? Why or why not? When the bus is making a turn, it is necessarily accelerating. Therefore, the inside of the bus constitutes a noninertial frame of reference. As viewed from inside the bus, a thrown softball will appear to be deflected by some mysterious external force. [Recall the video of the people in the rotating chamber – the same effect would occur inside the bus.] In actuality, the thrown softball is still following its typical parabolic path (as viewed from an observer in an inertial frame of reference).

Q4.28

5 points

A horse is hitched to a wagon. Since the wagon pulls back on the horse just as hard as the horse pulls on the wagon, why doesn’t the wagon remain in equilibrium, no matter how hard the horse pulls? It is true that the force exerted on the wagon by the horse is equal in magnitude and opposite in direction to the force exerted on the horse by the wagon (Newton’s third law). These two forces, however, do not cancel one another, because they are exerted on different objects (the first force is exerted on the wagon while the second force is exerted on the wagon).

Q4.33

8 points

Consider a tug-of-war between two people who pull in opposite directions on the ends of a rope. By Newton’s third law, the force that A exerts on B is just as great as the force that B exerts on A. So what determines who wins? (Hint: Draw a free-body diagram showing all the forces that act on each person.) Each contestant in the tug-of-war competition will experience four forces: (1) a downward gravititional force exerted by the Earth which is balanced by (2) an upward contact force exerted by the surface of the ground, (3) a horizontal force exerted by the rope (in the direction of the other contestant), which is opposed by (4) a horizontal frictional force exerted by the surface of the ground (in the direction opposite of the other contestant). For two contestants of equal mass, we conclude that whichever contestant experiences a larger frictional force with the ground will win the contest.

4.10

6 points

A dockworker applies a constant horizontal force on 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s? (a) Using ∆x = v0 t + 21 at2 and solving for a we find that a= Using Newton’s second law

P

2∆x 2(11.0 m) 2 = = 0.88 m/s . t2 (5.00 s)2

F = ma and solving for m we find that P F 80.0 N m= = 90.9 kg. = a 0.88 m/s2

(b) At the end of the 5.00 s, the block has a velocity 2

v = v0 + at = (0.88 m/s )(5.00 s) = 4.4 m/s. Therefore, if the worker stops pushing at the end of 5.00 s, the block will slide a distance of ∆x = vt = (4.4 m/s)(5.00 s) = 22 m over the next 5.00 s.

4.24

6 points

The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration? The reaction forces are: (1) The elevator passenger exerts a downward force of magnitude 620 N on the floor, and (2) The elevator passenger exerts an upward force of magnitude 650 N on the Earth. The passenger experiences a net downward force of 30 N. Therefore, the passenger is also accelerating in the downward direction. The mass of the elevator passenger is m=

650 N mg = 66.3 kg. = g 9.8 m/s2

Therefore, the magnitude of the passenger’s downward acceleration is P F 30 N 2 a= = = 0.45 m/s . m 66.3 kg 4.43

10 points

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope. A woman wearing golf shoes (so she can get traction on the ice) pulls horizontally on theh 6.00-kg crate with a force F that gives the crate an acceleration of 2.50 m/s2 . (a) What is the acceleration of the 4.00-kg crate? (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton’s second law to find the tension T in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, force T or force F ? (d) Use part (c) and Newton’s second law to calculate the magnitude of the force F . (a) Because the rope that connects the two crates is assumed to be of fixed length, the 4.00-kg crate has the same acceleration (namely, 2.50 m/s2 ) as the 6.00-kg crate. (b) The forces acting on the 4.00-kg crate are: (1) a downward gravitational force of magnitude 39.2 N, (2) an upward (contact) force of magnitude 39.2 N, and (3) a rightward force of magnitude T . Using Newton’s second law, we find that 2

T = (4.00 kg)(2.50 m/s ) = 10 N.

(c) The forces acting on the 6.00-kg crate are: (1) a downward gravitational force of magnitude 58.8 N, (2) an upward (contact) force of magnitude 58.8 N, and (3) a leftward force of magnitude T = 10 N, and (4) a rightward force of magnitude F . Since the 6.00-kg crate is accelerating to the right, the net force acting on it must also be to the right; therefore, force F has a larger magnitude than force T . (d ) Using Newton’s second law, we find that 2

F = (10 N) + (6.00 kg)(2.5 m/s ) = 25 N.

4.54

10 points

The two blocks in Fig. P4.54 are connected by a heavy uniform rope with a mass of 4.00 kg. An upward force of 200 N is applied as shown. (a) Draw three free-body diagrams: one for the 6.00-kg block, one for the 4.00-kg rope, and another one for the 5.00-kg block. For each force, indicate what body exerts that force. (b) What is the acceleration of the system? (c) What is the tension at the top of the heavy rope? (d) What is the tension at the midpoint of the rope? (a) The forces acting on the 6.00-kg block are: (1) an upward force of magnitude F = 200 N exerted by some unknown external agent, (2) a downward gravitational force of magnitude m1 g = 58.8 N exerted by the Earth, and (3) a downward force of magnitude T1 exerted by the top of rope. The forces exerted on the 5.00-kg block are (1) a downward gravitational force of magnitude m2 g = 49 N and (2) an upward force of magnitude T2 exerted by the bottom of the rope. The forces exerted on the 4.00-kg rope are (1) a downward gravitational force of magnitude mr g = 39.2 N, (2) an upward force of magnitude T1 exerted by the 6.00-kg block, and (3) a downward force of magnitude T2 exerted by the 5.00-kg block. (b) Treating the two blocks and the rope as our system, and using Newton’s second law, we find that a

200 N − (15.0 kg)(9.8 m/s2 ) F − (m1 + m2 + mr )g = m1 + m2 + mr 15.0 kg

=

2

= 3.53 m/s , in the upward direction. (c) Using Newton’s second law for the 6.00-kg block, we find that T1 = F − m1 (g + a) = (200 N) − (6.00 kg)(13.33 m/s2 ) = 120 N. (d ) Using Newton’s second law for the 5.00-kg block, we find that 2

T2 = m2 (g + a) = (5.00 kg)(13.33 m/s ) = 66.7 N. Therefore, the tension at the midpoint of the rope is T = (T1 + T2 )/2 = 93 N . Chapter 5 Q5.2

5 points

“In general, the normal force is not equal to the weight.” Give an example where these two forces are equal in magnitude, and at least two examples where they are not. If a block is resting on a horizontal surface, the normal force exerted on the block by the surface is equal in magnitude to the block’s weight. There are numerous examples, however, where these two forces are not equal in magnitude. Rememeber, normal forces and contact forces are the same thing – anytime you push on an object, you are exerting a normal force on that object (and vice versa!).

Q5.26

6 points

If you throw a baseball straight upward with speed v0 , how does it speed, when it return to the point from where you threw it, compare to v0 (a) in the absence of air resistance and (b) in the presence of air resistance? Explain. (a) In the absence of air resistance, the baseball experiences a constant acceleration (both while it is traveling upward and while it is traveling downward) of magnitude g. Thus, when it returns to the point from where you threw it, its speed is equal to the speed v0 with which you initially threw it. (b) In the presence of air resistance, the symmetry noted in part (a) is no longer present. While traveling upward, the magnitude of the baseball’s downward acceleration is greater than g. As it travels downward, the magnitude of the baseball’s downward acceleration is less than g. As a result of this asymmetry, when it returns to the point from where you threw it, its speed will be less than the speed v0 with which you initially threw it.

5.22

12 points

A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of neglible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) = At + Bt2 , where A and B are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50 m/s2 and 1.00 s later an upward velocity of 2.00 m/s. (a) Determine A and B, including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel? (a) The rocket’s acceleration is a(t) =

d dv At + Bt2 = A + 2Bt. = dt dt

At t = 0, a = 1.50 m/s2 , which means that A = 1.50 m/s2 . At t = 1.00 s, v = 2.00 m/s, which means that B = 0.5 m/s3 . (b) At 4.00 s, the rocket has an acceleration of a(t = 4.00 s) = (1.50 m/s2 ) + 2(0.5 m/s3 )(4.00 s) = 5.50 m/s2 .

(c) Using Newton’s second law, and solving for the thrust F , we find that (at t = 4.00 s) 2 F = m(a + g) = (2540 kg) 15.3 m/s = 3.89 × 104 N. This force is 1.56 times the weight of the rocket. (d ) At t = 0, a = 1.50 m/s2 . Therefore, the initial thrust due to the fuel is 2 F = m(a + g) = (2540 kg) 11.3 m/s = 2.87 × 104 N, which is 1.56 times the weight of the rocket. 5.36

9 points

A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As the angle α is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp? (a) Using the problem we did in class as an example [see lecture notes for Section 5.3], we find that the minimum angle at which the box starts to slip is α = tan−1 µs = tan−1 (0.35) = 19.3◦ .

(b) Using Newton’s second law, we find that once the box has begun to move, its acceleration is 2

2

a = (sin α − µk cos α) g = (sin 19.3◦ − (0.25) cos 19.3◦ ) (9.8 m/s ) = 0.927 m/s . (c) Using v 2 = v02 + 2a∆x, we find that after sliding 5.0 m along the ramp, the box will be moving h i1/2 2 v = 2(0.927 m/s )(5.0 m) = 3.04 m/s.

5.44

6 points

A flat (unbanked) curve on a highway has a radius of 220.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of friction that will prevent sliding? (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely? (a) Friction between the tires and the pavement is the force responsible that allows the car to round the curve. Since, in this case, the car is undergoing uniform ciruclar motion, the frictional force acting on the tires must be equal in magnitude to fs = mv 2 /r. If we suppose that the car’s tires are on the verge of sliding on the pavement, then fs = µs N = µs mg. Therefore, µs =

(25.0 m/s)2 v2 = 0.29. = 2 gr (9.8 m/s )(220.0 m)

(b) If the coefficient of friction between the tires and pavement √ is only one-third this value, then, by the above expression, the maximum speed must be 1/ 3 the value use in part (a); namely, vmax = 14.4 m/s . 5.47

6 points

In another version of the “Giant Swing”, the seat is connected to two cables as shown in Fig. E5.47, one onf which is horizontal. The seat swings in a horizontal circle at a rate of 32.0 rpm (rev/min). If the seat weighs 255 N and an 825-N person is sitting in it, find the tension in each cable. The combined weight of the seat and the person (1080 N) must be balanced by the upward component of the tension in the top cable. Therefore, the top cable has a tension T1 =

1080 N = 1410 N. cos 40.0◦

The horizontal component of the top cable’s tension, along with the tension in the lower cable, are responsible for the swing’s circular motion, so T1 sin θ + T2 = mv 2 /R = 4π 2 mR/T 2 where m = 110 kg is the combined mass of the seat and the person and T = 1.875 s is the swing’s period. Therefore, the tension in the lower cable is T2

= =

5.72

6 points

4πmR 4π 2 (110 kg)(7.50 m) − T sin θ = − (1410 N) sin 40.0◦ 1 T2 (1.875 s)2 8360 N.

Block A in Fig. P5.72 weighs 60.0 N. The coefficient of static friction between the block and the surface on which it rests is 0.25. The weight w is 12.0 N and the system is in equilibrium. (a) Find the friction force exerted on block A. (b) Find the maximum weight w for which the system will remain in equilibrium. (a) Since the system is in equilibrium, the tension in the lower cable is w. By the symmetry of the problem, that means that the tension in the cable connected to block A is also w. Thus, the frictional force exerted on block A has a magnitude w = 12.0 N. (b) The maximum weight w for which the system will remain in equilibrium is w = µs NA = (0.25)(60.0 N) = 15.0 N.

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