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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Laws of Motion

JIPMER 2019 · Physics · Laws of Motion

The rate of decrease of kinetic energy is 9.6 J/s. Find the...

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 m/s3 \mathrm{~m} / \mathrm{s}.

Options

  1. A3.2 N
  2. B4.8 N
  3. C2.4 N
  4. D5.6 N

Correct answer

A. 3.2 N

Step-by-step solution

Given, rate of decrease in kinetic energy dkdt=9.6 J/s\frac{\mathrm{dk}}{\mathrm{dt}}=9.6 \mathrm{~J} / \mathrm{s} When, speed, v=3 m/s\mathrm{v}=3 \mathrm{~m} / \mathrm{s} then force F=\mathrm{F}= ? Kinetic energy k=12mv2\mathrm{k}=\frac{1}{2} \mathrm{mv}^2 differentiating with respect to ' tt ', we have dkdt=12 m⋅2vdvdt=v⋅mdvdt=v.mdpdtdkdt=v.F[∵ force F=mdpdt]∴F=1v⋅dkdt=13×9.6=3.2 N\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}
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