to push a 25 kg crate up a 27 incline, a worker exerts a force of 120 N, parallel to the incline. As the crate slides 3.6 m, how much work is done on the crate by a) the worker b) force of gravity c) normal force

Answer 1

Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate.

Fc = 245N @ 27deg.
Fp = 245sin27 = 111.2N.=Force parallel to incline.
Fv = 245cos27 = 218.3N. = Force perpendicular to incline = Normal.

a. W = Fd = 120 * 3.6 = 432J.

b. W = Fc*d=Fc*h=245 * 3.6sin27 = 400J.

c. W=Fv * d = 218.3 * 3.6sin27 = 357J.

Answer 2

i would like to say thank you for this wonderful problem .. this same problem appeared at my homework and it was a big help

Answer 3

OMG! thankful! ♥️