t o a he ig h t of 8 m over a t ime o f 4 s. Th e fo rkl ift t he n ho l ds th e crate i n pl ace fo r 2 0 s.- .How much power does the forklift exert in lifting the crate?
(A)0 W
(B)2.0 103 W
(C)3.2 103 W
(D)2.0 104 W
(E)3.2 104 W - .How much power does the forklift exert in holding the crate in place?
(A)0 W
(B)400 W
(C)1.6 103 W
(D)4.0 103 W
(E)1.6 104 W
Explanations
- C
When the force is exerted in the direction of motion, work is simply the product of force and displacement.
The work done is ( 10 N)(4.0 m) = 40 J.
- D
The work done on the box is the force exerted multiplied by the box’s displacement. Since the box travels at
a constant velocity, we know that the net force acting on the box is zero. That means that the force of the
person’s push is equal and opposite to the force of friction. The force of friction is given by , where is
the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the box,
which is mg = ( 10 kg )( 10 m/s^2 ) = 100 N. With all this in mind, we can solve for the work done on the box:
- C
The work done by the force of gravity is the dot product of the displacement of the box and the force of
gravity acting on the box. That means that we need to calculate the component of the force of gravity that is
parallel to the incline. This is mg sin 30 = ( 10 kg)( 10 m/s^2 ) sin 30. Thus, the work done is