During a rockslide, a 330 kg rock slides from rest down a hillside that is 640 m

Question

During a rockslide, a 330 kg rock slides from rest down a hillside that is 640 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.20. If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide is 9.70×10^5 J. The mechanical energy that is dissipated by the frictional force during the slide is 3.66×10^5 J. What is the kinetic energy of the rock as it reaches the bottom of the hill, and its speed?

Explanation / Answer

Use conservation of energy, so (9.70x105 J)-(3.66x105 J)=6.04x105 J. This is equal to the kinetic energy at the bottom of the hill, so 6.04x105 J=1/2mv2 and then we plug in the value for the mass and solve for v to get v=60.5 m/s

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