Energy Balance

The power dissipated is the difference between the power supplied by the motor and the rate at which mechanical energy is stored in the sand.

1. Power Supplied ($P_{in}$):

From Question 22, $P_{in} = 2 \mu g h$.

2. Rate of Energy Storage ($P_{stored}$):

This includes the rate of increase of Kinetic Energy and Potential Energy.

• Rate of KE: $\frac{dK}{dt} = \frac{1}{2} (\frac{dm}{dt}) v^2 = \frac{1}{2} \mu v^2$.
  Substituting $v^2 = gh$: $\frac{dK}{dt} = 0.5 \mu g h$.
• Rate of PE: The sand is lifted to height $h$ at a rate of $\mu$.
  $\frac{dU}{dt} = (\frac{dm}{dt}) g h = \mu g h$.

$$ P_{stored} = 0.5 \mu g h + \mu g h = 1.5 \mu g h $$

3. Power Dissipated ($P_{diss}$):

$$ P_{diss} = P_{in} – P_{stored} $$ $$ P_{diss} = 2 \mu g h – 1.5 \mu g h = 0.5 \mu g h $$

Correct Option: (c) $0.5 \mu g h$