Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let $h$ be the height that the beam was lifted initially. The work done by the motor is equal to the increase in potential energy. That is: $W_1 = mgh$ We can write an expression for the power in the initial case. $P = \frac{W_1}{t}$ Let's suppose that the beam is lifted to twice the height. We can find an expression for the work done by the motor. $W_2 = mg~(2h) = 2~W_1$ We can write an expression for the power in the second case. $P_2 = \frac{W_2}{t/2}$ $P_2 = \frac{4~W_1}{t}$ $P_2 = 4~P$ The power increases by a factor of 4.