MCQ
Force acting on a particle moving in a straigth line varies with the velocity of the particle as $F =\frac {K}{v},$ where $K$ is a constant. The work done by this force in time $t$ is
- A$\frac{K}{{{v^2}}}t$
- B$2Kt$
- ✓$Kt$
- D$\frac {2Kt}{v^2}$
✓
Answer
Correct option: C.
$Kt$
c
$\mathrm{F}=\frac{\mathrm{k}}{\mathrm{v}}$
$\mathrm{F}=\frac{\mathrm{k}}{\mathrm{v}}$
$\mathrm{W}=\int \mathrm{F} \mathrm{dx}=\int \mathrm{k} / \mathrm{v} \mathrm{dx}\left(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\right)$
$=\int \frac{\mathrm{k}}{\frac{\mathrm{dx}}{\mathrm{dt}}} \mathrm{dx}=\int \mathrm{F} \cdot \mathrm{dt}$
$\mathrm{W}=\mathrm{Kt}$
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