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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2024 · 31 January · Evening Shift · Physics · Work, Power and Energy

A body of mass 2kg begins to move under the action of a time...

A body of mass 2kg2kg begins to move under the action of a time dependent force given by \overset{\to}{F} = \left(6t\overset{^}{i} + 6{t}^{2}\overset{^}{j}\right)N. The power developed by the force at the time tt is given by:

Options

  1. A(6t4+9t5)W\left(6{t}^{4} + 9{t}^{5}\right)W
  2. B(3t3+6t5)W\left(3{t}^{3} + 6{t}^{5}\right)W
  3. C(9t5+6t3)W\left(9{t}^{5} + 6{t}^{3}\right)W
  4. D(9t3+6t5)W\left(9{t}^{3} + 6{t}^{5}\right)W

Correct answer

D. (9t3+6t5)W\left(9{t}^{3} + 6{t}^{5}\right)W

Step-by-step solution

Given: \overset{\to}{F} = \left(6t\overset{^}{i} + 6{t}^{2}\overset{^}{j}\right)N

Now, \overset{\to}{F} = m\overset{\to}{a} = \left(6t\overset{^}{i} + 6{t}^{2}\overset{^}{j}\right)

⇒ \overset{\to}{a} = \overset{\to}{F}m = \left(3t\overset{^}{i} + 3{t}^{2}\overset{^}{j}\right)

Hence, velocity \overset{\to}{v} = \mathop{∫}_{0}^{t}\overset{\to}{a}dt = \dfrac{3{t}^{2}}{2}\overset{^}{i} + {t}^{3}\overset{^}{j}

Required power, P=F→.v→=(9t3+6t5)WP = \overset{\to}{F} . \overset{\to}{v} = \left(9{t}^{3} + 6{t}^{5}\right)W

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