JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A body of mass \(m\) starts moving from rest along \(x-\) axis so that its velocity varies as \(v = a\sqrt s\) where \(a\) is a constant and \(s\) is the distance covered by the body. The total work done by all the forces acting on the body in the first second after the start of the motion is
- A \(\frac{1}{8}\,m{a^4}{t^2}\)
- B \(4\,m{a^4}{t^2}\)
- C \(8\,m{a^4}{t^2}\)
- D \(\frac{1}{4}\,m{a^4}{t^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{8}\,m{a^4}{t^2}\)
Step-by-step Solution
Detailed explanation
From qusestion, \(v = a\sqrt s = \frac{{ds}}{{dt}}\) or, \(\begin{gathered} 2\sqrt s \,\,\, = at \Rightarrow s = \frac{{{a^2}{t^2}}}{4} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,F = m \times \frac{{{a^2}}}{2} \hfill \\ \end{gathered} \) Work done…
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