Question

To push a 26.0 kg crate up a frictionless incline, angled at 25.00 to the horizontal, a worker exerts a force of 214 N, parallel to the incline. Assume that the crate slides 1.50 m. (a) What is the work done on the crate by the worker's applied force? (b) What is the work done by the gravitational force on the crate? (c) What is the work done by the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

Answer 1

(a)W = F*s = 214*1.5 =321 N-m

(b) W= -mgsin $\theta$ * s = - 26*9.8*1.5*sin 25 = - 161.52 N-m (negative sign because crate moves against pull of gravity)

(c) Normal force is at 90 degree to the displacement, so it does zero work

W = 0 J

(d) Total work done = 321 - 161.52 = 159.48 N-m