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Physics Problem?
A 45 kg trunk is pushed 5.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is 0.20.
a) Calculate the work done by the applied horizontal force.
b) Calculate the work done by the weight of the trunk.
c) How much energy was dissipated by the frictional force acting on the trunk?
1 Answer
- NCSLv 75 years agoFavorite Answer
This is a little tricky because the applied force has a component normal to the incline.
a) normal force Fn = mgcosΘ + FsinΘ
Fn = 45kg * 9.8m/s² * cos30º + Fsin30º = 382N + 0.5F
so the friction force f = µ*Fn = 0.20*(382N + 0.5F) = 76.4N + 0.1F
For "constant speed," the sum of the upslope forces must equal the sum of the downslope forces:
FcosΘ = mgsinΘ + f
Fcos30º = 45kg * 9.8m/s² * sin30º + 76.4N + 0.1F = 297N + 0.1F
0.766F = 297 N
F = 388 N
The component of this force acts along 5.8 m, so
work = 388N * cos30º * 5.8m = 1947 J
b) The weight acts against the vertical displacement:
work = -mgdsinΘ = -45kg * 9.8m/s² * 5.8m * sin30º = -1279 J
c) "constant speed" means no increase in KE, so
friction work = 1947J - 1279J = 668 J
Check:
f = 76.4N + 0.1*388N = 115 N, so
friction work = -115N * 5.8m = 668 J √√√
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