JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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
- A \(\frac{{2F}}{{\sqrt {mk} }}\)
- B \(\frac{F}{{\pi \sqrt {mk} }}\)
- C \(\frac{{\pi F}}{{\sqrt {mk} }}\)
- D \(\frac{F}{{\sqrt {mk} }}\)
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 \(\,{W_{sp}}\) + \(w_f\) = \(\Delta K.E\)…
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