SOLUTION SET 2

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.

2. 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

3. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place 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?
If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration
Solution:

a. same
b. more
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.

4. 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.

5. 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?
Solution: vf = 0.0 m/s; vi = 4.0 m/s. Ff = m (mg). Using vf2 – vi2 = 2 a x, where a = m g, we re-write this as vf2 – vi2 = 2 m g x. We are looking for “x.” And, since vf = 0.0 m/s, we can re-write this as
– vi2 = 2 m g x, and dividing both sides by (2 m g), we get (– vi2)/(2 m g) = x; thus,
x = [-(4.0)2 /(2)(0.46)(-9.8)] = [-16/-9.016] = 1.77 meters, or 1.8 m.

6. 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.
Solution: We can use our “favorite” relationship, again: vf2 – vi2 = 2 a x, where the runner starts from rest, so vi = 0, thus changing the relationship to vf2 = 2 a x; when we re-write this, we get (vf2)/(2x) = a or, a = (13)2/(2)(22 m) = (169)/(44) = 3.84 m/s2. Therefore, (a/g) = coefficient of friction, or m = a/g = (3.84 m/s2)/(9.8 m/s2) = 0.39.

7. 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?
Solution:
Same
Same

8. 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.
Solution: Each point is 30° away from the vertical down, so the contact points will result on each of them having ½ m g x cos 30°, or F = (0.5)(.15kg)(9.8)(0.866) = 0.6365 N, or 0.64 N.

9. 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?
Solution:
No
Yes
Yes
No

10. 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.
Solution: Zero, as the net distance from start to a new cycle is zero.

11. 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.
Solution: Work = Force x Distance. Here, we have distance, but the “Force” here is really the “net force in the same direction,” And what is this net force? Fnet = F0Cos 55°.= (50.0 N)(0.573576436) = 28.67882182 N.
a. So, now, W = (28.67882182 N)(0.5 meter) =14.3 J.
b. Here, Fnet = F0Cos 65°.= (50.0 N)(0.4226) = 21.1 N, so, W = (21.1 N)(0.5 m) = 10.6 J.

12. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution: Work = Force x distance. We don't have the force, yet. But, Force = m a. We have the mass, but not the acceleration. Acceleration, a = Dv/Dt. We have Dv, but not Dt. To accelerate from rest means vi = 0.0 m/s and since vf = 7.7 m/s, we then have Dv = 7.7 m/s. Since we cannot get the force or time change, we have to try to solve this another way. For example, work also equals a change in kinetic energy, i.e., W = DKE = KEf – KEI = ½ m vf2 - ½ mvi2 = ½ m vf2 only, since vi = 0.0 m/s.
In the end, W = ½ m vf2 = (0.5)(73 kg) (7.7 m/s)2 = (0.5)(73 kg) (59.29) = 2164 J = 2.2 x 10^3 J.

13. 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?
Solution: An object in “free fall” without any air resistance would fall 16 meters in a time, “t”. But what is “t”? Using the relationship h = ½ g t2, we can re-write this as t = √[(2h)/(g)] = √[(2)(16)/(9.8)] = √[(32)/(9.8)] = √(3.27) = 1.81 seconds. And the final velocity in such a free fall would be vf = g t = (9.8)(1.81) = 17.7 m/s. Since the true landing speed is only 13 m/s, then we must say that = anet t where “anet” is the net acceleration, or vf = 13 m/s = anet (1.81 s). Thus, anet = (13)/(1.81) = 7.2 m/s2 . Thus, the force of air resistance, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.
a. W = Fair x d = (0.36 N)(16 m) = 5.8 J.
b. We found this already, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.

14. 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.
Solution: We see that the speed goes from 19 to 12, or, 7.0 m/s over the 32 meters. Would it go from 19 to 15.5 over 16 meters? Using vf2 – vi2 = 2 a x, we see that the deceleration, a, will be (vf2 – vi2)/(2x) = (144 – 361)/(32) = (-217)/(64) = - 3.39 m/s2. We assume that it's a constant deceleration. If so, then if we plug in 16 for x instead of 32, we will find out if the final velocity is 15.5 or not: using vf2 – vi2 = 2 a x and re-writing as vf2 = vi2 + 2 a x where a = - 3.39 and this time, x = 16, we get: (19)2 + (2)(-3.39)(16) = 361 – 108 = 253, or, vf = 15.9 m/s. The answer is NO, but it is very close.
b. we did that already, vf = 15.9 m/s.

15. 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.
Solution: W = ½ k x2, so, k = 2 W / (x2) = (2)(180) / (0.15)2 = (360)/(0.0225) = 16,000 N/m.

16. 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.
Solution: Same amount of energy for all three, but more lost for farther distances traveled.