Question

A 0.075 kg ball in a kinetic sculpture is raised 1.33 m above the ground by a motorized vertical conveyor belt. A constant frictional force of 0.350 N acts in the direction opposite the conveyor belts motion. How much total work is done in raising the ball?

Answer 1

Answer : The total work done in raising the ball is, 0.98 J

Explanation : Given,

Mass of the ball = 0.075 kg

Height raised of the ball = 1.33 m

As we know that the object is moving with the constant velocity, that means the work done against the gravity will be the net-work done.

So, the work done will be:

$w=mgh$

where,

w = work done

m = mass of ball

h = height of ball

g = acceleration due of gravity = $9.8m/s^2$

Now put all the given values in the above formula, we get:

$w=(0.075kg)\times (9.8m/s^2)\times (1.33m)$

$w=0.97755J=0.98J$

Thus, the total work done in raising the ball is, 0.98 J