Question

A fully loaded, slow-moving freight elevator has a cab with a total mass of $1200\text{kg}$, which is required to travel upward by $54\text{m}$ in $3\text{min}$, starting and ending at rest. The elevator’s counterweight has a mass of only $950\text{kg}$, so the elevator motor must help. What average power is supplied by force which the motor exerts on the cab via the cable? [Take $g=10{\text{m/s}}^2$]

• A. $750\text{W}$
• B. $650\text{W}$
• C. $550\text{W}$
• D. $450\text{W}$

Answer 1

The correct option is A $750\text{W}$

On applying work-energy theorem,
$W_{\text{all forces}}=\Delta KE$
$\Rightarrow W_{\text{gravity}}+W_{\text{motor}}=\Delta KE$

[only forces acting are by gravity and the motor]

$\Rightarrow -1200\times 10\times 54+950\times 10\times 54+W_{motor}=0$

[since the counterweight descends by the same distance over which the elevator moves up]

$\Rightarrow W_{motor}=135000\text{J}$

So, power supplied by motor
$P=\frac{W_{motor}}{t}=\frac{135000}{3\times 60}$
$\Rightarrow P=750\text{W}$