Difference between revisions of "Sample problem page"
From TuHSPhysicsWiki
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:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 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) | :(Since the velocity is constant, there is no acceleration so they are weightless) | ||
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===Problem 6. === | ===Problem 6. === | ||
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===Problem 7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?=== | ===Problem 7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?=== | ||
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:so now we can use Newton's second law | :so now we can use Newton's second law | ||
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N | :F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N | ||
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===Problem 8. === | ===Problem 8. === | ||
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===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> 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? === | ===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> 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? === | ||
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:22 = 9.8m + 4.5m = 14.3m | :22 = 9.8m + 4.5m = 14.3m | ||
:m = 22/14.3 = 1.538 kg = 1.5 kg | :m = 22/14.3 = 1.538 kg = 1.5 kg | ||
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===Problem 10. === | ===Problem 10. === | ||
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Use v2 = u2 + 2as, a = 30.1786 m/s/s | Use v2 = u2 + 2as, a = 30.1786 m/s/s | ||
F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N | F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N | ||
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===Problem 12. === | ===Problem 12. === | ||
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===Problem 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? === | ===Problem 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? === | ||
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<63 N - 98 N> = (10 kg) a | <63 N - 98 N> = (10 kg) a | ||
a = -3.5 m/s/s (down) | a = -3.5 m/s/s (down) | ||
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15. A 75-kg petty thief wants to escape form 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. | 15. A 75-kg petty thief wants to escape form 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. | ||
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<568.4 N - 735 N> = (75 kg)a | <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 | a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s | ||
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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? | 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? | ||
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<21750 N - 20580 N> = (2100 kg) a | <21750 N - 20580 N> = (2100 kg) a | ||
a = +.557 m/s/s = +.56 m/s/s upward | a = +.557 m/s/s = +.56 m/s/s upward | ||
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Revision as of 15:07, 30 August 2007
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
- 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 Problem 6.
- 1.7 Problem 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 Problem 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 Problem 10.
- 1.11 Problem 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 Problem 12.
- 1.13 Problem 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?
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
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
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)
Problem 6.
Problem 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
Problem 8.
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
Problem 10.
Problem 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 Table of Contents
Problem 12.
Problem 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) Table of Contents
15. A 75-kg petty thief wants to escape form 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 Table of Contents
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 Table of Contents
