Difference between revisions of "Sample problem page"
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| + | [[Main Page]] > [[User:Cmurray]] > [[Murray's Sand Box]] > Sample problem page <br /> <br /> | ||
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== Problems == | == Problems == | ||
| − | ===1 | + | ===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: | :Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so: | ||
:F = ma | :F = ma | ||
| Line 9: | Line 10: | ||
---- | ---- | ||
| − | ===2=== | + | ===2.=== |
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | ===3 | + | |
| − | + | ===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: | + | :They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> 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 | :F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N | ||
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | ===4=== | + | |
| + | ===4. === | ||
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | === | + | |
| + | ===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: | :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 earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N | ||
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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) | ||
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | === | + | |
| + | ===6. === | ||
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | === | + | ===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: | :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 | ::90 km/hr/3.6 = 25 m/s | ||
| Line 44: | Line 53: | ||
: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 | ||
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | === | + | ===8. === |
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
===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? === | ||
| Line 53: | Line 64: | ||
: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 | ||
| + | [[#top | Table of Contents]] | ||
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| − | === | + | ===10. === |
| − | + | [[#top | Table of Contents]] | |
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| − | === | + | ===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: | + | :We have a cute linear kinematics problem to solve first: |
| − | s = 2.8 m | + | :s = 2.8 m |
| − | u = 0 (assumed) | + | :u = 0 (assumed) |
| − | v = 13 m/s | + | :v = 13 m/s |
| − | a = ??? | + | :a = ??? |
| − | t = Don't care | + | :t = Don't care |
| − | 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 |
| + | [[#top | Table of Contents]] | ||
---- | ---- | ||
| − | === | + | ===12. === |
| + | [[#top | Table of Contents]] | ||
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| − | === | + | ===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: | + | :First calculate the weight: |
| − | F = ma = (10 kg)(9.80 m/s/s) = 98 N | + | :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: | + | :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 | + | :<63 N - 98 N> = (10 kg) a |
| − | a = -3.5 m/s/s (down) | + | :a = -3.5 m/s/s (down) |
| − | + | [[#top | Table of Contents]] | |
| + | ---- | ||
| + | ===14. === | ||
| + | [[#top | Table of Contents]] | ||
| + | ---- | ||
| + | ===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 | ||
| + | [[#top | Table of Contents]] | ||
| + | ---- | ||
| − | + | ===16. === | |
| − | + | [[#top | 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 | |
| − | + | [[#top | Table of Contents]] | |
| − | + | ---- | |
| − | 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 | ||
| − | |||
Latest revision as of 11:44, 3 March 2009
Main Page > User:Cmurray > Murray's Sand Box > Sample problem page
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?
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)
6.
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.
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.
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.
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.
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.
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
