During a rockslide, a 690 kg rock slides from rest down a hillside that is 740 m

Question

During a rockslide, a 690 kg rock slides from rest down a hillside that is 740 m along the slope and 360 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.34. (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?

Explanation / Answer

Part A)

U = mgh

U = (690)(9.8)(360)

U = 2434320 J   which is 2.43 X 106 J

Part B)

The thermal energy will be the work done by friction

W = Ffd = mgcosd

The angle of the incline is found by sin-1 (360/740)

= 29.1o

W = (.34)(690)(9.8)(cos 29.1)(740)

W = 1486565 J which is 1.49 X 106 J

Part C)

KE = U - W

KE = (2434320 - 1486565) = 947755 J   which is 9.48 X 105 J

Part D)

KE = .5mv2

947755 = (.5)(690)(v2)

v = 52.4 m/s

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