Work Done by the Gravitational Force problem

In summary, the textbook drops to a friend's outstretched hands at distance D = 10.0 m below. The gravitational force does 166.5 J of work on the book as it falls to her hands. The potential energy of the book-earth system is 196 J when the textbook is released. If the gravitational potential energy of that system is zero at ground level, its potential energy U when the textbook reaches the hands is 29.4 J.
  • #1
kappcity06
90
0
this proplem has been giving me a lot of trouble. I was wondering if anyone could lead me in the right direction.

You drop a 2.00 kg textbook to a friend who stands on the ground at distance D = 10.0 m below. Your friend's outstretched hands are at distance d = 1.50 m above the ground.

a)How much work Wg does the gravitational force do on the book as it drops to her hands? (i got this answer-166.5J)

b)What is the change U in the gravitational potential energy of the textbook-Earth system during the drop? (I could not find a way to get this answer.)

c)If the gravitational potential energy of that system is zero at ground level, what is its potential energy U when the textbook is released?(196J)

d)If the gravitational potential energy of that system is zero at ground level, what is its potential energy U when the textbook reaches the hands? (29.4J)

e) How much work Wg is done on the textbook by its weight as it drops to your friend's hands if U is 100 J at the ground level?(166.5)

f) What is the change U in the gravitational potential energy of the textbook-Earth system during the drop if U is 100 J at the ground level?

g) Find U at the release point when U is 100 J at the ground level?(296J)

h) Find U at the hands when U is 100 J at the ground level? (129.4)

so i need help finding b and f. please and thankyou in advance
 
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  • #2
I don't understand how you could do c, d, g, or h, without knowing how to calculate the change in gravitational potential energy near the Earth's surface. (Which is all you need for b and f.)
 
  • #3
You appear to be hung up on the term "textbook-earth system".

Is the potential energy of the book-earth system any different than the potential energy of the book in the Earth's gravitational field? You have already calculated the latter.

AM
 
  • #4
If you lift a book a distance h, on the surface of the earth, then you have increased the book's gravitational potential energy (which is how I would interpret "gravitational potential energy of the textbook-Earth system") by gh. That is assuming that h is small enough that the acceleration due to gravity can be taken to be the constant -h.
 
  • #5
i'm still uin sure of the answer. I did the height(10m)*gravity(9.8)*mass(2) and the answer is wrong?
 
  • #6
kappcity06 said:
I did the height(10m)*gravity(9.8)*mass(2) and the answer is wrong?
The book drops onto his hand, not onto the ground.
 
  • #7
kappcity06 said:
i'm still uin sure of the answer. I did the height(10m)*gravity(9.8)*mass(2) and the answer is wrong?

Keep in mind that the potential energy decreases as the book falls.

You're on the right track
 
  • #8
remember:

[tex]\{Delta} U = mgh_{2} - mgh_{1}[/tex]

with 2 = situation after movement and 1 = situation before movement.
 
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  • #9
would it be mass*gravity*height(10m)-mass*gravity*height(1.5m)
 
  • #10
would it be mass*gravity*height(10m)-mass*gravity*height(1.5m)

this did not work. I'm still getting a wrong answer. i have 166.6.
 
  • #11
i got part b which was -166.6. but part f is still giving be trouble. should i just and 100 to the answer for b
 
  • #12
Part f still asks for the change in gravitational PE.
 
  • #13
kappcity06 said:
i got part b which was -166.6. but part f is still giving be trouble. should i just and 100 to the answer for b

so you were only looking for the minus sign :rolleyes:

if you did b correct, than f should be no problem.
 
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  • #14
i tried adding one hundred to get -66.6 but that was wrong. should i subtract 100 to get -266.6
 
  • #15
Hint: The 100 J is irrelevant in finding the change in PE.

(It's like asking what's the difference in height from your head to your toes if you are standing on Mt. Everest. Do you need to know the height of Everest to answer?)
 
  • #16
thank you doc al the answer was the same as b. thanks to all who helped me too
 

FAQ: Work Done by the Gravitational Force problem

What is work done by the gravitational force?

The work done by the gravitational force is the amount of energy transferred when an object is moved by the force of gravity. It is calculated by multiplying the force of gravity by the distance an object is moved in the direction of the force.

How is work done by the gravitational force different from potential energy?

Work done by the gravitational force is the energy transferred when an object is moved, while potential energy is the energy an object possesses due to its position or configuration. The two are related, as the work done by the gravitational force can change an object's potential energy.

What is the formula for calculating work done by the gravitational force?

The formula for work done by the gravitational force is W = Fd, where W is work, F is the force of gravity, and d is the distance an object is moved in the direction of the force.

Can work done by the gravitational force be negative?

Yes, work done by the gravitational force can be negative if the force and movement are in opposite directions. This means that the gravitational force is actually doing work on the object, causing it to lose potential energy.

What factors affect the work done by the gravitational force?

The work done by the gravitational force is affected by the mass of the object, the strength of the gravitational force, and the distance the object is moved. The work done also depends on the direction of the force and movement, with opposite directions resulting in negative work.

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