A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.

Question

A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.52 m/s. The pulling force is 106 N parallel to the incline, which makes an angle of 19.4 degree with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.06 m. How much work is done by the gravitational force on the crate? Determine the increase in internal energy of the crate-incline system owing to friction. How much work is done by the 106-N force on the crate? What is the change in kinetic energy of the crate? What is the speed of the crate after being pulled 5.06 m?

Explanation / Answer

(b)Internal energy due to friction = work done against friction = umgCos19.4*(distance)
= 0.4*10.6*9.81*Cos19.4*(5.5 ) = 198.5 J
(d) Energy remain conserved
initial kinetic energy = final kinetic energy + potential energy+ friction energy
Initial - final = potential + friction = mg(5.5Sin19.4)+198.5 = 373.27 J

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