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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 sec. will be :


  • Option 1)

     4.5 J


  • Option 2)

    22 J


  • Option 3)

     9 J


  • Option 4)

     18 J


Answers (1)


As we discussed in

Work done by time dependent force -

W= \int \vec{F}\cdot \vec{v}\: dt

- wherein

Where \vec{F} and \vec{v} are force and velocity vector at any instant


 Force is given F = 6t

F = ma = m\frac{dv}{dt}= 6t

\frac{dv}{dt}=6t(Since m = 1 kg)

dv = 6t dt

On integrating \int_{0}^{v}dV =6\int_{0}^{1}t dt = 3

Therefore v = 3 m/s

Therefore change in kinetic energy in one second  = \frac{1}{2}\times m \times 3^{2}-0=4.5 J


Option 1)

 4.5 J


This is the correct option.

Option 2)

22 J


This is an incorrect option.

Option 3)

 9 J


This is an incorrect option.

Option 4)

 18 J

This is an incorrect option.

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