JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2019
- Hard
- MCQ
A block of mass \(m\), lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant \(k\). The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force \(F\), the maximum speed of the block is

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Answer & Solution
Correct Answer
(D) \(\frac{F}{{\sqrt {mk} }}\)
Step-by-step Solution
Detailed explanation
When \(\,{v_{\max }}\) \(\Rightarrow \) acceleration \(= 0\)
\( \Rightarrow \,x = \frac{F}{K}\)
Apply work energy theorem
\(\,{W_{sp}}\) + \(w_f\) = \(\Delta K.E\)
\(\begin{array}{l}
- \frac{1}{2}K{x^2} + F.x = \Delta K.E\,\,\,;\,\, - \frac{1}{2}K\frac{{{F^2}}}{{{K^2}}} + \frac{{{F^2}}}{K}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}mu_{\max }^2\\
\frac{{{F^2}}}{{2K}} = \frac{1}{2}mu_{\max }^{2\,}\,\,\,;\,\,\,\frac{F}{{\sqrt {mK} }} = {V_{\max }}
\end{array}\)
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