Physics · Work, Power & Energy

JEE Main 2025 — 7 April, Morning Shift — Question 48

An object of mass 1000 g experiences a time dependent force F⃗=(2ti^+3t2j^)N\vec{F}=\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{N}. The power generated by the force at time tt is:

  1. Option A:

    (2t2+3t3)W\left(2 t^{2}+3 t^{3}\right) \mathrm{W}

  2. Option B:

    (3t3+5t5)W\left(3 t^{3}+5 t^{5}\right) \mathrm{W}

  3. Option C:

    (2t2+18t3)W\left(2 t^{2}+18 t^{3}\right) \mathrm{W}

  4. Option D:

    (2t3+3t5)W\left(2 t^{3}+3 t^{5}\right) \mathrm{W}

    Correct

Answer: D

Step-by-step solution

m=1000m=1000 gram F⃗=(2ti^+3t2j^)\vec{F}=\left(2 t \hat{i}+3 t^{2} \hat{j}\right)

So, dvdt=Fm=(2ti^+3t2j^)\frac{d v}{d t}=\frac{F}{m}=\left(2 t \hat{i}+3 t^{2} \hat{j}\right)

So ∫0vdv=∫0t(2ti^+3t2j^)dt\int_{0}^{v} d v=\int_{0}^{t}\left(2 t \hat{i}+3 t^{2} \hat{j}\right) d t

So v=(ti^+t3j^)\quad v=\left(t \hat{i}+t^{3} \hat{j}\right) So power =F⃗⋅V⃗=2t2+3t5=\vec{F} \cdot \vec{V}=2 t^{2}+3 t^{5}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Work, Power & Energy
Topic
Power