Question

Ball with mass .5kg is released from rest at point A, 5m above the bottom of the tank, it falls through the tank of oil and at point B it is 2m above the bottom with speed of 4m/s. What is the work done by Force of friction between A and B?

Answer 1

The friction did 10.7 J of work.

Explanation:

The ball descended 3 m (from 5 m to 2 m). So it lost gravitational potential energy

`GPE_"lost" = m*g*h = 0.5 kg*9.8 m/s^2*3 m = 14.7 J`

Some of those 14.7 J were converted to kinetic energy

`KE = 1/2 *m*v^2 = 1/2*0.5 kg*(4 m/s)^2 = 4 J`

So 4 J of the original 14.7 J were converted to kinetic energy. The remainder of those 14.7 J were lost to the work done by the friction with the oil. So to calculate the amount of work done

`GPE_"lost" = KE + "work"`

`14.7 J = 4 J + "work"`

`"work" = 14.7 J - 4 J = 10.7 J`

Those 10.7 J heated the oil.

I hope this helps,
Steve