Chapter 5: Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143
1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, How long does it take for the object to accelerate from rest to a speed of “v” now?
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.
3. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
Solution: In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi//h in a distance of 1.7 meters.
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution: m, v, x; F?
98 mi/h ~ 160 km/h = 160,000 m/3600s = 44.4 m/s
F = m a = (0.15 kg)(1162 m/s2) = 174 N.
a = v/t = 98/t; t = v/a = x/v
v = x/t; t = x/v; v2 = a x
a = v2/x = (44.4)2 / 1.7 m = 1975/1.7 = 1162 m/s2.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters.
b. Describe how you came to that solution.
Solution: vi, m, vf, x, F? Vi = 202 mph ~ 320 km/h = 320,000 m/h = 320,000/3600 m/s = 88.9 m/s, and 45 mph ~ 72 km/h = 72,000 m/h = 72,000 m/3600 sec = 20 m/s.
We know that F = m a, and we have m, but must find “a.” Using vf2 – vi2 = 2 a x, we see that a = (vf2 – vi2)/2x = [(20)2 – (88.9)2]/(2)(185) = [400-7901]/(370) = - 20.27 m/s2. Thus, F = (891)(20.27 m/s2) = - 18064 N = - 1.8 x 10^3 N.
19. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They pace their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.
25. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction)?
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, starting with the smallest. Indicate ties where appropriate.
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
Solution: Each arm has a y-component of (290 N)(Cos 24) = 264.93 N, so two arms has ~530 N. The 67-kg gymnast would normally have m g = (67)(9.8) = 657 N, so the difference is (657 - 530) = 127 N.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During he storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184
1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. The the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
3. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
9. A tie of uniform width is laid out ion a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
13. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
19. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
39. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
53. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be a the North Pole? Explain.
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213
1. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
13. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
25. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
29. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248
1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
3. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
19. Suppose the situation descrige in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth. Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.,
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Solution: skip this one
Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292
1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”
The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e.,
p = m v.
In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).
pcar = 15,800 kg-m/s and pball = (0.142 kg)(vball), so now we need to find vball.
vball = (15,800 kg-m/s) / (0.142 kg) = 111,268 m/s = 111 km/sec = 69.54 miles/sec = 250,352 mi/hour = 2.5 x 105 mi/hr.
3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.
Solution: Dog is 26.2 kg, 2.7 m/s @90°, cat 5.3 kg, 3.04 m/s @ 0°. Homosapien, 74 kg.
p1 = m1v1 = (26.2)(2.7) = 70.7 kgm/s @ 90°
p2 = m2v2 = (5.3)(3.04) = 16.1 kgm/s @ 0°.
pt2 = (70.7)2 + (16.1)2 = 4998.5 + 259.6 = 5258
pt = (5258)½ = 72.5 m/s;
16.1 = 72.5 cos a
cos a = (16.1)/(72.5) = 0.222
a = cos-1 (0.222) = 77°
p = 72.5 kgm/s @ 77°.
v = p/m = (72.5 kgm/s)/(74.0 kg) = 0.98 m/s = 9.8 x 10-1 m/s @ 77°.
9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
Solution: Momentum, p = m v, but it is equivalent to F Dt.
(a) for the boulder, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
for the pebble, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
Thus, the same
13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
Solution: Impulse is force multiplied by the small incremental change in time, I = F Dt, so in this case, I = (1250.00 N)(5.95 x 10-3 s) = 74337.5 10-3 = 74.3375 = 7.43 x 101 N-s.
19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
Solution: Remember that I = F Dt = D(mv) = mDv. (The mass does not change).
(a) I = (0.14 kg)(- 36 – 18) = (0.14 kg)(– 54) = - 7.5 N-s.
(b) It would double.
(c) no change.
25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
Solution: In this case, there is no momentum in the system, so after the astronaut pushes the satellite, the net momentum is still zero. So, m1v1 + m2v2 = 0. But m1 = 92 kg, v1 = ?; m2 = 1200 kg, v2 = - 0.14 m/s. So, let's set up a ratio type thing:
m1v1 + m2v2 = 0
m1v1 = - m2v2
v1 = - [(m2) / (m1)] v2 = - [(1200 kg) / (92 kg)] (- 0.14 m/s) = + 1.83 m/s. If the astronaut traveled 1.83 m/s for 7.5 seconds, then he traveled a distance of x = (1.83)(7.5)= 13.7 m, or 1.37 x 101 meters.
29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast.
Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
Solution: In a “closed” system, which this is, we don't have any outside sources of speed, momentum, force, or acceleration. So, the momentum of both vehicles before the impact must equal the momentum of both vehicles (stuck together) after the collision.
pi = pc + pv = mcvc + mvvv = pf = (mc + mv)vf .And we are looking for vv.
mcvc = (950 kg)(20 m/s) = 19,000 kg-m/sec @ 0°. This is the initial momentum for the car.
mvvv = (1300 kg)(21 m/s) = 27,300 kg-m/sec @ 90°. This is the initial momentum for the van.
As the initial momenta are at right angles, then the net original momentum would be the vector addition of pc + pv = 19,000 @ 0° + 27,300 @ 90°. In this case, using Pythagoras, a = 19,000 and b = 27,300, so we need c, which we get from the Pythagorean relationship: a2 + b2 = c2, or, (1.9 x 104)2 + (2.73 x 104)2 = (c)2 ; = (3.61 x 108) + (7.45 x 108) = 11.06 x 108. Therefore, c = √(11.06 x 108) = 3.33 x 104 kg-m/s. But what is the angle, θ?
Tanθ = (27300)/(19000) = (27.3)/(19) = 1.437; so the angle is: θ = tan-1 (1.437) = 55.17°.
So, in the end, the final momentum of the wreckage is 3.33 x 104 kg-m/s @ 55.17°
However, what if the angle is 40.0°? Then that means that Tanθ = tan 40.0° = 0.839. So, pv/pc = 0.839; and pc is still 19,000 kg-m/s @ 0°. Thus, pv/pc = (pv)/(19,000) = 0.839. Or, pv = (19000)(0.839) = 15,943 kg-m/s. Since the mass of the van is 1300 kg, then the initial velocity of the mini van is: pv = mvvv = (1300 kg)(vv) = 15,943 kg-m/s, or, vv = (15,943 kg-m/s) / (1300 kg) = 12.26 m/s.
31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”
The answer is vf = 5.0 m/s.
In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
Solution: Okay, the relationship for kinetic energy is: KE = ½ m v2.
(a) The KE of the car, or KE1 = ½ m1 v12 = (0.5)(1200)(2.5)2 = 3,750 Joules.
The KE of the truck, or KE2 = ½ m2 v22 = (0.5)(2600)(6.2)2 = 49,972 Joules.
So, KE1 + KE2 = 3750 Joules + 49,972 Joules = 53,722 Joules. But KEf = ½ (m1 + m2)vf2 = (0.5)(3800 kg)(5.0 m/s)2 = 47,500 Joules. Since 53,722 Joules > 47,500 Joules, then KEf < KE1 + KE2 or KE1 + KE2 > KEf.
(b) we did this already in part (a)
39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
(a) 4.55 + 7.81 = 12.36 m/s.
(b) The hefalump gave it more energy.
53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
Solution: This is the same as if we had a wooden stick of 1 meter whose center of mass is at 0.5 meters in the y-direction, and a wooden stick of 2 meters, whose center of mass is at 1.0 meters in the x-direction.
(a) So the center of mass is some where between (0, 0.5) and (1.0, 0). And in fact, it's at
(b) no change