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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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