During a rockslide, a 380 kg rock slides from rest down a hillside that is 740 m

Question

During a rockslide, a 380 kg rock slides from rest down a hillside that is 740 m along the slope and 290 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.19. (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

a)potential energy at any height h=mass*g*h

so at 290 m height, potential energy=380*9.8*290=1.08*10^6 J

b) energy transferred to thermal energy=work done against friction

=friction force*distance travelled

angle of the hill side with horizontal=arctan(height/length)=arctan(290/740)=21.4 degree

then normal force=mass*g*cos(21.4)

=380*9.8*cos(21.4)

=3467.3 N

then friction force=friction coefficient*normal force

=0.19*3467.3

=658.78 N

now distance travelled from the top to bottom=length of the incline=740 m

hence work done against friction force=friction force*distance travelled

=658.78*740=4.875*10^5 J

part c:

hence kinetic energy left with the rock=initial potential energy-work done against friction

==>kinetic energy=1.08*10^6-4.875*10^5=592500 J

part d:

as kinetic energy=0.5*mass*speed^2

==>speed=sqrt(2*kinetic energy/mass)

=sqrt(2*592500/380)

=55.843 m/s

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