JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 ............... \(\mathrm{J}\)
- A \(4.5\)
- B \(22\)
- C \(9\)
- D \(18\)
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,…
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