A 284 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surface is 0.513. How much power must the engine deliver to move the crate at a constant speed of 7.2 m/s?

Answer 1

M*g = 284kg * 9.8N./kg = 2783.2 N. = Wt. of the crate = Normal force(Fn).

Fk = u*Fn = 0.513 * 2783.2 = 1427.8 N. =
Force of kinetic friction.

Fap-Fk = M*a
Fap-1427.8 = M*0 = 0
Fap = 1427.8 N. = Force applied.

Power = Fap * V = 1427.8 * 7.2 = 10,280
J/s. = 10,280 Watts.