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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2017
  • Hard
  • MCQ

A time dependent force \(F = 6t\) acts on a particle of mass \(1\ kg\). If the particle starts from rest, the work done by the force during the first \(1\) secand will be ............... \(\mathrm{J}\)

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Answer & Solution

Correct Answer

(A) \(4.5\) 

Step-by-step Solution

Detailed explanation

Using, \(F=ma = m\frac{{dV}}{{dt}}\)

\(6t = 1.\frac{{dV}}{{dt}}\)  [\(\because m = 1\; kg\) given]

\(\int\limits_0^v {dV = \int {6t\,dt} } \)

\(V = 6\left[ {\frac{{{t^2}}}{2}} \right]_0^1\, = 3\,m{s^{ - 1}}\,\,\,\)  [\(\because  t=1 \;sec\) given]

Form woek-energy theorem,

\(W = \Delta KE = \frac{1}{2}m\left( {{V^2} - {u^2}} \right) = \frac{1}{2} \times 1 \times 9 = 4.5\,J\)

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