Posted by **Anonymous** on Tuesday, September 27, 2011 at 7:09pm.

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

- Physics -
**Henry**, Thursday, September 29, 2011 at 3:25pm
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.

- Physics -
**Angat Vora**, Wednesday, March 28, 2012 at 7:50pm
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)

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