Physics
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
J
(b) the increase in internal energy in the boxfloor system due to friction
J
(c) the work done by the normal force
J
(d) the work done by the gravitational force
J
(e) the change in kinetic energy of the box
J
(f) the final speed of the box
m/s

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
(f) KE=mv²/2
v= sqrt(2•KE/m)