Question

A force of constant magnitude pushes 6.5-kg box 3.4 m up a vertical surface, as shown in the figure. The coefficient of friction is 0.35 and the angle θ is 30.0°. a) Determine the work done on the box by F. b) Determine the work done on the box by the force of gravity. c) Determine the work done on the box by the normal force. d) Determine the increase in gravitational potential energy of the system of the box and Earth as the box moves up the wall.

Answer 1

a)

net force mg + frictional force = F* sin 30

mg = F* ( sin 30 - 0.35* cos 30 )

F = mg/(sin30 - 0.35*cos30) = 6.5*9.8/( sin 30 - 0.35* cos 30 )

F = 323.529 N

work done on the box W = F*S = 323.529*3.4

W = 1099.998 J  

b)

the work done on the box by the force of gravity W = mgh

W = 6.5*(-9.8)*3.4 = -216.58 J

c)

the work done on the box by the normal force W = 0 J

d)

increase in gravitational potential energy E = mgh = 6.5*9.8*3.4

PE = 216.58 J