Giancoli Physics (5th ed) Chapter 2
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Contents
- 1 Problems
- 1.1 1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?
- 1.2 2.
- 1.3 3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?
- 1.4 4.
- 1.5 5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?
- 1.6 6.
- 1.7 7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?
- 1.8 8.
- 1.9 9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s2 using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?
- 1.10 10.
- 1.11 11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s?
- 1.12 12.
- 1.13 13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down?
- 1.14 14.
- 1.15 15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.
- 1.16 16.
- 1.17 17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking?
- 1.18 18.
- 1.19 19.
- 1.20 20.
- 1.21 21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?
- 1.22 22.
- 1.23 23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?
- 1.24 24.
- 1.25 25.
- 1.26 26.
- 1.27 27.
- 1.28 28.
- 1.29 29.
- 1.30 30.
- 1.31 31.
- 1.32 32.
- 1.33 33.
- 1.34 34.
- 1.35 35.
- 1.36 36.
- 1.37 37.
- 1.38 38.
- 1.39 39.
- 1.40 40.
- 1.41 41.
- 1.42 42.
- 1.43 43.
- 1.44 44.
- 1.45 45.
- 1.46 46.
- 1.47 47.
- 1.48 48.
- 1.49 49.
- 1.50 50.
- 1.51 51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?
- 1.52 52.
- 1.53 53.
- 1.54 54.
- 1.55 55.
- 1.56 56.
- 1.57 57.
- 1.58 58.
- 1.59 59.
- 1.60 60.
- 1.61 61.
- 1.62 62.
- 1.63 63.
- 1.64 64.
- 1.65 65.
- 1.66 66.
- 1.67 67.
- 1.68 68.
- 1.69 69.
- 1.70 70.
- 1.71 71.
- 1.72 72.
- 1.73 73.
Problems
1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?
- Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:
- F = ma
- F = (60.0 kg)(1.15 m/s/s) = 69 N
2.
- (70.8 km/h)
3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?
- They give us m = 9.0 g = 9.0 x 10-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:
- F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N
4.
- (105 km/h, 29 m/s, 95 ft/s)
5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?
- Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:
- on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N
- on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N
- on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N
- At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N
- (Since the velocity is constant, there is no acceleration so they are weightless)
6.
- (about 300 m/s)
7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?
- First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:
- 90 km/hr/3.6 = 25 m/s
- We have a cute linear kinematics problem to solve:
- s = don't care
- u = 90 km/hr/3.6 = 25 m/s
- v = 0
- a = ???
- t = 8.0 s
- Use v = u + at to find a:
- v = u + at
- 0 = 25 m/s + a(8.0 s)
- a = -3.125 m/s/s
- so now we can use Newton's second law
- F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N
8.
- (10.4 m/s, +3.5 m/s)
9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s2 using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?
- The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:
- <22 N - mg> = m(+4.5 m/s/s)
- <22 - 9.8m> = m(4.5)
- 22 = 9.8m + 4.5m = 14.3m
- m = 22/14.3 = 1.538 kg = 1.5 kg
10.
- (4.43 h, 881 km/h)
11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s?
- We have a cute linear kinematics problem to solve first:
- s = 2.8 m
- u = 0 (assumed)
- v = 13 m/s
- a = ???
- t = Don't care
- Use v2 = u2 + 2as, a = 30.1786 m/s/s
- F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N
12.
- (6.73 m/s)
13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down?
- First calculate the weight:
- F = ma = (10 kg)(9.80 m/s/s) = 98 N
- Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:
- <63 N - 98 N> = (10 kg) a
- a = -3.5 m/s/s (down)
14.
- (5.2 s)
15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.
- If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.
- Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:
- <568.4 N - 735 N> = (75 kg)a
- a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s
16.
- (-6.3ms-2, .64 g)
17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking?
- First calculate the weight:
- F = ma = (2100 kg)(9.80 m/s/s) = 20580 N
- Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:
- <21750 N - 20580 N> = (2100 kg) a
- a = +.557 m/s/s = +.56 m/s/s upward
18.
19.
20.
21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?
- -The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.
- -A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:
- (30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s
- The initial velocity here is zero, so we can drop it from the equation.
- 900= 2(3.0 m/s/s) s
- s= 1.5x102 m
22.
23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?
- -To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.
- -We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.
- a= (25 m/s) / (5 sec)
- a= - 5 m/s/s, because it is slowing down
- -Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.
- (0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s
- 0= 625 + (-10) s
- -625= (-10) s
- s= 62.5 m
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51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?
- -(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s
- -(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.
- -(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s
- -(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s
- -(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s
