posted by Vince on .
A 39.0 kg box initially at rest is pushed 4.50 m along a rough, horizontal floor with a constant applied horizontal force of 135 N. If the coefficient of friction between box and floor is 0.300, find the following.
(a) the work done by the applied force
(b) the increase in internal energy in the box-floor system due to friction
(c) the work done by the normal force
(d) the work done by the gravitational force
(e) the change in kinetic energy of the box
(f) the final speed of the box
m=39.0 kg s=4.5 m,
F=135 N, μ=0.3
(a) W(F) = Fs
(b) ΔU= W(fr) = μmgs
(c) W(N) = 0
(d) W(mg) = 0
(e) KE=W(F) –W(fr) =Fs - μmgs