Question

A man is pulling a block resting on the ground by applying force $F=(20t+40)\text{N}$ as shown in the figure. If the pulley is frictionless and mass of the block is $4\text{kg}$, find out the power delivered by the force, $3\text{seconds}$ after the man starts pulling the block. (Take $g=10{\text{m/s}}^2$)

• A. $2250\text{J/s}$
• B. $5250\text{J/s}$
• C. $3000\text{J/s}$
• D. $1250\text{J/s}$

Answer 1

The correct option is A $2250\text{J/s}$

Tension in the string = Applied force ($F$)
Given,
$F=20t+40\text{N}$
Total force on the block $=F-mg$
$⟹$ $ma=20t+40-mg$
$⟹4a=20t+40-4\times 10$
$⟹4a=20t$
$⟹a=5t$ (acceleration of the block)
But $a=\frac{dv}{dt}$
$⟹5t=\frac{dv}{dt}$

Integrating on both sides:
${\int }_0^{v}dv={\int }_0^{3}5tdt$
$⟹v=\frac{5t^2}{2}{|}_0^3$
$⟹v=2.5\times 9=22.5\text{m/s}$

Force${|}_{t=3\text{s}}=20\times 3+40=100\text{N}$
Therefore, Power at $t=3\text{s}$
$=F.v=100\times 22.5=2250\text{J/s}$