During a rockslide, a 500 kg rock slides from rest down a hillside that is 750 m

Question

During a rockslide, a 500 kg rock slides from rest down a hillside that is 750 m along the slope and 340 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.29. (a) If the gravitational potential energy U of the rock-Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then? Formulas used would be nice.

Explanation / Answer

a) U = mgh = 1666 KJ,,,,,,,,,,,, b) energy transferred to thermal energy during the slide = 0.29*500gh= 483.14 KJ,,,,,,, c)KE at bottom= 1182.86 KJ,,,,,,,,,,, d) v=68.78 m/s

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