An electric motor raises a load of $0.2\ kg$, at a constant speed, through a vertical distance of $3.0\ m$ in $2\ s$. If the acceleration of free fall is $10\ m/s^2$, the power in $W$ developed by the motor in raising the load is:
(a) 0.3
(b) 1.2
(c) 3.0
(d) 6.0

Mass raised by the electric motor $m=0.2\ kg$

Aceleration of free fall $g=10\ m/s^2$

Height $h=3\ m$

Time taken to raise the load $t=2\ s$

Therefore, work done by the motor to raise the load $=mgh$

$=0.2\ kg\times10\ m/s^2\times3\ m$

$=6.0\ J$

Therefore, power$=\frac{work\ done}{time}$

$=\frac{6.0\ J}{2\ s}$

$=3.0\ Watt$

So, option (c) is correct.

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Updated on: 10-Oct-2022

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