posted by amy .
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
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.
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