Question

A 0.28-kg stone is held 1.5 m above the top edge of a water well and then dropped into it. The well has a depth of 4.8 m.
(a) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system before the stone is released?
J

(b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system when it reaches the bottom of the well?
J

(c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?
J

Answer 1

a. PE = mgh = 0.28 * 9.8 *1.5 = 4.1J.

b. PE = 0.28 * 9.8 * 4.8 = 13.2 Joules.

c. Change = mg(h1+h2) - 0 =
0.28 * 9.8 * (1.5+4.8) - 0 = 17.3J.

Answer 2

Actually,

** GPE = mgh

a) GPE = mgh
as height of well is taken to be area of 0 GPE.
sign is positive.
GPE = mgh = 0.28*9.8*1.5


b) GPE = mgh
as height is now below area of 0 GPE, h changes to -h.
therefore sign is negative
GPE = 0.28*9.8*(-4.8)

c) GPE = mgh
here, mg(-h1-h2)
therefore,
GPE = 0.28*9.8*(-1.5-4.8)