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posted by  CHMSTRY on 11/6/2009 11:15:47 PM  |  status: Closed  |  Earned Karma: 55

physics

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/10/2009 at 10:00:00 AM
Question Details:
Dropping a Textbook You drop a 1.60 kg textbook to a friend who stands on the ground 10.0 m below the textbook with outstretched hands 1.50 m above the ground (Fig. 10-26).


Figure 10-26

(a) How much work Wgrav is done on the textbook by the gravitational force as it drops to your friend's hands?
1 J
(b) What is the change ΔU in the gravitational potential energy of the textbook-Earth system during the drop?
2 J
(c) If the gravitational potential energy U of that system is taken to be zero at ground level, what is U when the textbook is released?
3 J
(d) If the gravitational potential energy U of that system is taken to be zero at ground level, what is U when the textbook reaches the hands?
4 J
(e) How much work Wgrav is done on the textbook by the gravitational force as it drops to your friend's hands if U is 100 J at the ground level.
5 J
(f) What is the change ΔU in the gravitational potential energy of the textbook-Earth system during the drop if U is 100 J at the ground level.
6 J
(g) If the gravitational potential energy U of that system is taken to be 100 J at ground level, what is U when the textbook is released?
7 J
(h) If the gravitational potential energy U of that system is taken to be 100 J at ground level, what is U when the textbook reaches the hands?
8 J
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AAnswers:

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Sage
Karma Points: 3,983
posted by Mr. SinghSoorma on 11/7/2009 1:16:18 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
a) W = Force x distance
In this case, the force = mg
W = mg * d = (1.6)(9.8)(10-1.5) = 133.28J

b) ΔU  = Ub - Ua , where b is in the person's hands
ΔU  = mghf - mghi = mg (hf -hi) = (1.6)(9.8)(1.5-10) = -133.28

c) Utop = mgh = (1.6)(9.8)(10) = 156.8 J

d) Ubottom = mgh = (1.6)(9.8)(1.5) = 23.52 J

e) Work is still force x distance, so taking the answer from part a..133.28J...

f) W = - ΔU = - 133.28J ( change in potential energy is the same as if the ground still had PE = 0, only dependent on what the final and initial point are..in other words, the change in height is still the same)

g)156.8 J  + 100 J = 256. 8 J  ( add 100 J to the normal PE, when ground is 0)

h)23.52 J + 100 J = 123.52 J ( add 100J to normal PE, when ground is just 0 )
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