During a rockslide, a 480 kg rock slides from rest down a hillside that is 500 m

Question

During a rockslide, a 480 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.13.
(a) 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?
J

(b) How much energy is transferred to thermal energy during the slide?
J

(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
J

(d) What is its speed then?
m/s

Explanation / Answer

1)U=mgh=480kg*9.8m/s^2*300m=1.4e6 J.
2)Thermal Energy=m*g*cos()*distance*=480kg*9.8m/s^2*cos(tan-1 (300/500))*.13*sqrt(300^2+500^2)=2.11e5J

3)K=U-E=1.4e6-2.11e5=1.2e6J

4)K=1/2*m*v^2=.5*480kg*v^2=1.2e6 so v=70 m/s. Hope that helped!

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