Question

A body of mass $5\text{kg}$ starts from rest and moves in a straight line under the influence of two variable forces $F_1=(10t+\frac{6}{t})\text{N}$ and $F_2=(15-\frac{6}{t})\text{N}$, acting in same direction. Find the power delivered by force $F_1$ at $t=2\text{sec}$.

• A. $130\text{J/s}$
• B. $230\text{J/s}$
• C. $30\text{J/s}$
• D. $350\text{J/s}$

Answer 1

The correct option is B $230\text{J/s}$

Total force $F=F_1+F_2$
$F=(10t+\frac{6}{t})+(15-\frac{6}{t})$
$F=10t+15$
$⟹ma=10t=15$
$⟹5a=10t+15$
$⟹a=2t+3$
$⟹\frac{dv}{dt}=2t+3$

Integrating both sides:
${\int }_0^{v}dv={\int }_0^2(2t+3)dt$
$⟹v=(t^2+3t){|}_0^2$
$⟹v=2\times 2+3\times 2=10\text{m/s}$

Work done by force $F_1$
$=F_1.v=F_{1}v\cos \theta$
Here $\theta =0^{\circ }$
$\therefore$ Work done by force $F_1$ at $t=2\text{s}$
$=(10\times 2+\frac{6}{2})\times 10=230\text{J/s}$