A farmer uses a tractor to pull a 150 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 0.40. Part A What is the tractor's power output?

the tractor's power output = 1318.47 watts

Explanation:

Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons

Resolve that weight force into its components at right angles to the ramp and parallel to the slope.

Down the slope we have 1471.5sin (15°) = 380.85 N

At right angles to the ramp we have 1471.5cos (15°) = 1421.35 N

OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface. In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}

Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N

The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down

= (380 N + 568.54 N)

= 948.54N

We need the speed in m/s not km/h

5.0 km/h = 5000 m/h

= (5000/3600)

= 1.39 m/s

Power = (948.54 N * 1.39m/s)

= 1318.47 N. m/s

= 1318.47 watts