For a 21.1 kg rock is on the edge of a 96.2 m cliff what are the following:
The potential energy does the rock possess relative to the base of the cliff. If the rock falls from the cliff, what is its kinetic energy just before it strikes the ground and what speed does the rock have as it strikes the ground.
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The gravitational potential energy of a body placed at a height h is given by PE = m*g*h where m is the mass of the body and g is the acceleration due to gravity.
The mass of the rock is 21.1 kg and it rests on a cliff that is 96.2 m high. The gravitational potential energy of the rock relative to the base of the cliff is 21.1*9.8*96.2 = 19892.23 J
When the rock falls, just as it reaches the base all the potential energy has been converted to kinetic energy and this is equal to 19892.23 J.
An object with a mass m traveling at a velocity v has a kinetic energy of (1/2)*m*v^2. For the rock, the kinetic energy at the bottom of the cliff is 19892.23 J. If its speed is V
(1/2)*21.1*V^2 = 19892.23
=> V = sqrt (1885.52)
=> V = 43.42 m/s.
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