Q97 / 180 · JIPMER 2019 · Physics

The rate of decrease of kinetic energy is 9.6 J/s. Find the magnitude of force acting on particle when it's speed is $3 \mathrm{~m} / \mathrm{s}$.

  • Correct answer
    A
    3.2 N
  • B
    4.8 N
  • C
    2.4 N
  • D
    5.6 N

Solution

Given, rate of decrease in kinetic energy $\frac{\mathrm{dk}}{\mathrm{dt}}=9.6 \mathrm{~J} / \mathrm{s}$ When, speed, $\mathrm{v}=3 \mathrm{~m} / \mathrm{s}$ then force $\mathrm{F}=$ ? Kinetic energy $\mathrm{k}=\frac{1}{2} \mathrm{mv}^2$ differentiating with respect to ' $t$ ', we have $\begin{aligned} & \frac{\mathrm{dk}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~m} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v} \cdot \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}} \\ & =v . m \frac{\mathrm{dp}}{\mathrm{dt}} \\ & \frac{\mathrm{dk}}{\mathrm{dt}}=\mathrm{v} . \mathrm{F} \quad\left[\because \text { force } \mathrm{F}=\frac{\mathrm{mdp}}{\mathrm{dt}}\right] \therefore \quad \mathrm{F}=\frac{1}{\mathrm{v}} \cdot \frac{\mathrm{dk}}{\mathrm{dt}} \\ & =\frac{1}{3} \times 9.6=3.2 \mathrm{~N} \\ &\end{aligned}$

Hence, the correct answer is option: 'A'.

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