A 38.5 kg box initially at rest is pushed 5.80 m along a rough, horizontal floor
ID: 1606076 • Letter: A
Question
A 38.5 kg box initially at rest is pushed 5.80 m along a rough, horizontal floor with a constant applied horizontal force of 145 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 box-floor 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/sExplanation / Answer
a] Work done by the applied force = FS = 145(5.80) = 841 J
b] Increase in energy = work done by friction = umgS = 0.3(38.5)(9.8)(5.8) = 656.502 J
c] Work done by Normal force = RScos90 = 0 J
d] Work done by gravity = 0 J
e] change in kinetic energy of the box = Net Work done = 841 - 656.502 = 184.498 J
f] change in kinetic energy = (1/2)m[v2 - u2] = (1/2)m[v2 - 02]
=> v = 3.096 m/s.