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

From TuHSPhysicsWiki
Jump to: navigation, search
(New page: ==Problems== Return to Solutions ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== [[#top | Table of ...)
 
Line 1: Line 1:
 
==Problems==
 
==Problems==
 
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]
 
[[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]]
 
[[#top | Table of Contents]]
 
----
 
----
 +
 
===2.===
 
===2.===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===4.===
+
 
 +
===4. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===6.===
+
 
 +
===6. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===8.===
+
===8. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===10.===
+
 
 +
===10. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===12.===
+
===12. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===14.===
+
===14. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
===16.===
+
 
 +
===16. ===
 
[[#top | Table of Contents]]
 
[[#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]]
 
[[#top | Table of Contents]]
 
----
 
----
 +
 
===18.===
 
===18.===
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]

Revision as of 10:23, 5 September 2007

Contents

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

Table of Contents


2.

Table of Contents


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

Table of Contents


4.

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:
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)

Table of Contents


6.

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:
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

Table of Contents


8.

Table of Contents


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

Table of Contents


10.

Table of Contents


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


12.

Table of Contents


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


14.

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

Table of Contents


16.

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


18.

Table of Contents


19.

Table of Contents


20.

Table of Contents


21.

Table of Contents


22.

Table of Contents


23.

Table of Contents


24.

Table of Contents


25.

Table of Contents


26.

Table of Contents


27.

Table of Contents


28.

Table of Contents


29.

Table of Contents


30.

Table of Contents


31.

Table of Contents


32.

Table of Contents


33.

Table of Contents


34.

Table of Contents


35.

Table of Contents


36.

Table of Contents


37.

Table of Contents


38.

Table of Contents


39.

Table of Contents


40.

Table of Contents


41.

Table of Contents


42.

Table of Contents


43.

Table of Contents


44.

Table of Contents


45.

Table of Contents


46.

Table of Contents


47.

Table of Contents


48.

Table of Contents


49.

Table of Contents


50.

Table of Contents


51.

Table of Contents


52.

Table of Contents


53.

Table of Contents


54.

Table of Contents


55.

Table of Contents


56.

Table of Contents


57.

Table of Contents


58.

Table of Contents


59.

Table of Contents


60.

Table of Contents


61.

Table of Contents


62.

Table of Contents


63.

Table of Contents


64.

Table of Contents


65.

Table of Contents


66.

Table of Contents


67.

Table of Contents


68.

Table of Contents


69.

Table of Contents


70.

Table of Contents


71.

Table of Contents


72.

Table of Contents


73.

Table of Contents


74.

Table of Contents


75.

Table of Contents


76.

Table of Contents


77.

Table of Contents


78.

Table of Contents


79.

Table of Contents


80.

Table of Contents


81.

Table of Contents


82.

Table of Contents


83.

Table of Contents


84.

Table of Contents


85.

Table of Contents


86.

Table of Contents


87.

Table of Contents


88.

Table of Contents


89.

Table of Contents


90.

Table of Contents


91.

Table of Contents


92.

Table of Contents


93.

Table of Contents


94.

Table of Contents


95.

Table of Contents


96.

Table of Contents


97.

Table of Contents


98.

Table of Contents


99.

Table of Contents


100.

Table of Contents