You drop a 2 kg textbook to a friend who stands on the ground 10m below the testbook with outstreatched hands 1.5m above the ground. a) how much work W_g is done on the textbook by the gravitational force as it drops to your friend's hands?

Question

b)What is the change in the gravitational potential energy of the textbook -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 when the textbook c) is released and d) when it reaches the hands?

a) would I use W_g=mgh
=2(9.8)(10-1.5)
which gave me 194.5J

b) would IT be the exact same thing as a)?

For c) and d) I am not really sure what to do for these two problems.

Yes, it is the same, however, the gpe is reduced (negative sign).

For C, use initial height as 10 to find the GPE intial. For the final, use 1.5 as the h to get GPE final. The difference becomes Kinetic energy.

Answer 1

a) You used the correct formula for calculating the work done by the gravitational force. W_g = mgh = 2(9.8)(10 - 1.5) = 194.5 J.

b) Yes, the change in gravitational potential energy is the same as the work done by the gravitational force, but with a negative sign, as the potential energy is reduced during the drop. So, the change is -194.5 J.

c) To find the gravitational potential energy when the textbook is released, use the formula U = mgh. Here, the height is 10 m, so U_initial = 2(9.8)(10) = 196 J.

d) To find the gravitational potential energy when the textbook reaches the friend's hands, use the same formula with the height being 1.5 m. So, U_final = 2(9.8)(1.5) = 29.4 J.