A0.075 kg ball in a kinetic sculpture moves at a constant speed along a motorized vertical conveyor belt. the ball rises 1.32 m above the ground. a constant frictional force of 0.350 n acts in the direction opposite the conveyor belt's motion. what is the net work done on the ball?

0.51 J

Explanation:

The work done on the ball is equal to the sum of two terms:

- the work done in lifting the ball by 1.32 m

- the work done by the frictional force of 0.350 N that acts opposite to the motion; this work is negative, since the force acts in the opposite direction to the motion of the ball.

In formulas, the net work done is:

where

m = 0.075 kg is the ball's mass

g = 9.8 m/s^2 is the gravitational acceleration

h = 1.32 m is the height throug which the ball has been moved

Ff = 0.350 N is the frictional force

Substituting, we find

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle