A 0.22-kg stone is held 1.2 m above the top edge of a water well and then droppe

Question

A 0.22-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 5.1 m. (a) Relative to the configuration with the stone 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) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone?Earth 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

Explanation / Answer

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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