During a rockslide, a 770 kg rock slidesfrom rest down a hillside that is 500 m
ID: 1675215 • Letter: D
Question
During a rockslide, a 770 kg rock slidesfrom rest down a hillside that is 500 m long and 300 m high. Thecoefficient of kinetic friction between the rock and the hillsurface is 0.29. (a) If the gravitational potential energyU of the rock-Earth system is set to zero at the bottom ofthe hill, what is the value of U just before theslide?1 J
(b) How much energy is transferred to thermal energy during theslide?
2 J
(c) What is the kinetic energy of the rock as it reaches the bottomof the hill?
3 J
(d) What is its speed then?
4 m/s
Did you calculatethe initial kinetic energy and gravitational potential energy andthe final kinetic energy and gravitational potential energy? Howmuch energy is missing? (a) If the gravitational potential energyU of the rock-Earth system is set to zero at the bottom ofthe hill, what is the value of U just before theslide?
1 J
(b) How much energy is transferred to thermal energy during theslide?
2 J
(c) What is the kinetic energy of the rock as it reaches the bottomof the hill?
3 J
(d) What is its speed then?
4 m/s Did you calculatethe initial kinetic energy and gravitational potential energy andthe final kinetic energy and gravitational potential energy? Howmuch energy is missing?
Explanation / Answer
(a) The energy at the start is U = mgh m = 770 Kg, h = 300m, so U = 2.26x106 J (b) The force on the rock due to friction is F=N where N is the normal force and the coefficient offriction. The work done by this force is W = Fd where d is the distance. In this case d=500m The normal force is perp to the surface of the slope, and cancelsout the component of gravity in that direction. Hence N = mgcos where =sin-1[300/500] = 36.870 deg So W=dmgcos = 8.75x105 J (c) The rock loses this amount of energy, so the final energy is Ef = Ei - W Ef = 1.385x106 J (d) kinetic energy is T = (1/2)mv2 At the bottom of the slope, there is no potential energy (as theheight is zero), so all of Ef is kinetic energy. (1/2)mv2 = Ef v = 60.0 ms-1
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