A .20 kg stone is held 1.3 m above the top edge of a water well and then dropped

Question

A .20 kg stone is held 1.3 m above the top edge of a water well and then dropped into it. The well has a depth of 5.0 m. If the gravitational potential energy of the stone-Earth system is defined to be zero when the stone is at the top edge of the well, what is the gravitational potential energy of the stone-Earth system (a) just before the stone is dropped, and (b) just as it reaches the bottom of the well? (c) What is the change in the gravitational potential energy of the system from release to reaching the bottom of the well?

Explanation / Answer

Given Mass of the stone , m = 0.20 kg The depth of the wall , = 5.0 m Initially the ball is at the height, ho 1.3 m If the gravitational potential energy of the stone-
Earth system is defined as zero that is mghf = 0 a)The gravitational potential energy of the stone-Earth system just before the stone is dropped is, Wgr = mgho - mg hf        = 0.20 kg *9.8 m/s^2 *1.3 m        =2.54 J b)The gravitational potential energy of the stone-Earth system just as it reaches the bottom of the well, Wgr1 = mgh - mg hf        = 0.2kg*9.8 m/s^2(5m - 1.3m)
       = 7.25J C)Change in the the gravitational
potential energy of the system is ?Wgr = 7.25 J -2.54J          = 4.71J as it reaches the bottom of the well, Wgr1 = mgh - mg hf        = 0.2kg*9.8 m/s^2(5m - 1.3m)
       = 7.25J C)Change in the the gravitational
potential energy of the system is ?Wgr = 7.25 J -2.54J          = 4.71J

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