A rock of mass 500 kg slides down from rest
along a hill slope that is 500 m long and
300 m high and reaches the bottom. The
coefficient of kinetic friction, between the
rock and the hill slope is 0.25. Calculate
(i) the potential energy of the rock just
before sliding.
(ii) the work done on the rock by the
frictional force.
(iii) the speed of the rock at the bottom of
the hill,
Draw the free body diagram.
g=10ms2
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a) U = initial potential energy = M g h
= (520 kg)*(9.8 m/s^2)*300 m = 1,529,000 J
b) Energy transferred to heat = (friction force) x (500 m)
= M g * 0.25 * cos A* 500 = 509,600 J
The hill angle A has a cosine of 0.8
c) subtract (b) from (a) to get KE
d) use KE = (1/2) m V^2 and solve for V
a) U = initial potential energy = M g h
= (520 kg)*(9.8 m/s^2)*300 m = 1,529,000 J
b) Energy transferred to heat = (friction force) x (500 m)
= M g * 0.25 * cos A* 500 = 509,600 J
The hill angle A has a cosine of 0.8
c) subtract (b) from (a) to get KE
d) use KE = (1/2) m V^2 and solve for V
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