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 ............... \(\mathrm{J}\)
Pick an option to check your answer.
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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