Difference between revisions of "Giancoli Physics (5th ed) Chapter 2"

Revision as of 09:23, 5 September 2007

Problems

Return to Solutions

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

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2.

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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

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4.

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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)

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6.

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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

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8.

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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

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10.

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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:

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

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12.

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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:

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)

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14.

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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

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16.

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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?

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

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@@ Line 1: / Line 1: @@
 ==Problems==
 [[Giancoli Physics (5th ed) Solutions | Return to Solutions]]
-===1.===
+== 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
 [[#top | Table of Contents]]
 ----
 ===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<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
 [[#top | Table of Contents]]
 ----
-===4.===
+===4. ===
 [[#top | Table of Contents]]
 ----
-===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?===
+: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)
 [[#top | Table of Contents]]
 ----
-===6.===
+===6. ===
 [[#top | Table of Contents]]
 ----
-===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?===
+: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
 [[#top | Table of Contents]]
 ----
-===8.===
+===8. ===
 [[#top | Table of Contents]]
 ----
-===9.===
+===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? ===
+: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
 [[#top | Table of Contents]]
 ----
-===10.===
+===10. ===
 [[#top | Table of Contents]]
 ----
-===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? ===
+: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
 [[#top | Table of Contents]]
 ----
-===12.===
+===12. ===
 [[#top | Table of Contents]]
 ----
-===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? ===
+: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)
 [[#top | Table of Contents]]
 ----
-===14.===
+===14. ===
 [[#top | Table of Contents]]
 ----
-===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.===
+: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.===
+===16. ===
 [[#top | Table of Contents]]
 ----
-===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? ===
+: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]]
 ----
 ===18.===
 [[#top | Table of Contents]]

Difference between revisions of "Giancoli Physics (5th ed) Chapter 2"

Revision as of 09:23, 5 September 2007

Contents

Problems

Problems

1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?

2.

3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?

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?

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?

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?

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?

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?

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.

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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?

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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?