Question

For the problem below, I know how to work each part out but I want to know the reasoning behind parts (g) and (h). Why do you add the new U at y=0 to the potential energies found for parts (c) and (d)?

You drop a 2.80 kg book to a friend who stands on the ground at distance D = 11.0 m below. If your friend's outstretched hands are at distance d = 1.60 m above the ground (see the figure), (a) how much work Wg does the gravitational force do on the book as it drops to her hands? (b) What is the change ΔU in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy U of that system is taken to be zero at ground level, what is U (c) when the book is released and (d) when it reaches her hands? Now take U to be 100 J at ground level and again find (e) Wg, (f ) ΔU, (g) U at the release point, and (h) U at her hands. I know the answers for each of the problems are:

(a) =258, (b) =-158, (c) =302, (d) =43.9, (e) =258, (f)=-258 (g)=402, (h)=144

Answer #1

It's pretty simple. It's all about where you take the reference

The question straight away gives that potential energy (U) at ground level is already 100 J, so you have to add this 100 J of energy to find part ( g) and (h)

Okay, Let's take it in another way. Suppose there are two grounds ( one above the another). The potential energy of upper ground is 100 J with respect to lower ground. so when your friend ( standing on upper ground) catches the book or when you thrown a book, you have to add that 100 J of energy to get total potential of the system.

I hope you got the point

Let me know if you have any doubt.

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Question 110 pts
A 319 kg motorcycle is parked in a parking garage. If the car has
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units or the answer will be marked incorrect.
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