Question

A farmer uses a tractor to pull a 190 kg bale of hay up a 15∘ incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.45.

What is the tractor's power output?

Answer 1

Given
mass of bale is m = 190 kg , up the ramp of inclination theta= 15 degrees

and moving with a steady velocity 5.0 km/h by a tractor

coefficient of kinetic friction between the bale and the ramp is mue_k = 0.45

the tractor's power output P = ?

as the bale is moving on the ramp with friction and not accelerating , first find the forces acting on it

the forces along the ramp (downward) , mg sin theta and Frictional force (F_f)
the normal force F_N = mg cos theta and the applied force by the tractor (up the ramp )

the frictional force F_f = mue_k*F_N = mue_k*mg cos theta

so the net force is F = F_t - mg sin theta - F_f= 0

   F_t = mg sin theta + F_f

   F_t = mg sin theta + mue_k*mg cos theta
   F_t = mg (sin theta + mue_k*cos theta)
substituting the values

   F_t = 190*9.8(sin 15 +0.45 cos 15) N

   F_t = 1291.2703 N
speed v = 5.0 km/h = 5.0*5/18 m/s = 1.3889m/s

By definition power P = F*V = 1291.2703*1.3889 W = 1793.45 W