Question

During a rockslide, a 360 kg rock slides from rest down a hillside that is 500 m along the slope and 250 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.30. (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?

Answer #1

potential energy at the top of the hill = mass * g * h

potential energy = 360 * 9.81 * 250

**a) potential energy = 882900**
**J**

thermal energy = energy lost due to friction

energy lost due to friction = k * mg * cos(theta) * d

cos(theta) = sqrt(500^2 - 250^2) / 500

energy lost due to friction = 0.3 * 360 * 9.81 * (sqrt(500^2 - 250^2) / 500) * 500

energy lost due to friction = 458768.2974 J

**b) energy transferred to thermal energy =
458768.2974** **J**

kinetic energy at the bottom = total energy - energy lost

kinetic energy at the bottom = 882000 - 458768.2974

**c) kinetic energy at the bottom = 423231.7026**
**J**

kinetic energy = 0.5 * mv^2

kinetic energy = 0.5 * 360 * v^2

423231.7026 = 0.5 * 360 * v^2

v = 48.49 m/s

**speed = 48.49** **m/s**

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