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?

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.

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.

Yes, you are correct for part (a) in calculating the work done by the gravitational force. The equation W_g = mgh is used to calculate the work done by gravity where m is the mass (2 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the change in height (10 m - 1.5 m = 8.5 m). Plugging these values in, we get:

W_g = 2 kg * 9.8 m/s^2 * 8.5 m =
166.6 J

For part (b), the change in gravitational potential energy (ΔU) is equal to the negative of the work done by gravity. So, ΔU = -W_g = -166.6 J. Since the gravitational potential energy U is taken to be zero at ground level, the initial gravitational potential energy (U_i) when the textbook is released (c) is 0 J. The final gravitational potential energy (U_f) when the textbook reaches the hands (d) is equal to the negative of the change in gravitational potential energy, so U_f = -ΔU = 166.6 J.

a) Yes, you are correct in using the formula W_g = mgh to calculate the work done by the gravitational force on the textbook. Plugging in the values, we have:

W_g = (2 kg) * (9.8 m/s^2) * (10 m - 1.5 m)
W_g = 2 * 9.8 * 8.5
W_g = 164.6 J

So the work done by the gravitational force on the textbook as it drops to your friend's hands is 164.6 J.

b) You are right that the change in gravitational potential energy (ΔU) of the textbook-Earth system during the drop can be calculated using the same formula as in part a). However, there is a slight difference due to the reference level chosen for the zero potential energy.

Since the question states that the gravitational potential energy (U) is taken as zero at the ground level, we can calculate the initial and final gravitational potential energies relative to this reference point.

c) To calculate the gravitational potential energy (U) when the textbook is released, we can use the formula U = mgh. Plugging in the values, we have:

U_c = (2 kg) * (9.8 m/s^2) * (10 m)
U_c = 196 J

So when the textbook is released, the gravitational potential energy of the textbook-Earth system is 196 J.

d) Similarly, to calculate the gravitational potential energy (U) when the textbook reaches the hands, we can use the same formula. Plugging in the values, we have:

U_d = (2 kg) * (9.8 m/s^2) * (1.5 m)
U_d = 29.4 J

So when the textbook reaches the hands, the gravitational potential energy of the textbook-Earth system is 29.4 J.