JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A bead of mass ' \(m\) ' slides without friction on the wall of a vertical circular hoop of radius ' \(R\) ' as shown in figure. The bead moves under the combined action of gravity and a massless spring ( k ) attached to the bottom of the hoop. The equilibrium length of the spring is ' \(R\) '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' R ', would be (spring constant is ' k ', g is accleration due to gravity)
- A \(\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}\)
- B \(2 \sqrt{\mathrm{gR}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}\)
- C \(\sqrt{2 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}\)
- D \(\sqrt{2 \mathrm{Rg}+\frac{4 \mathrm{kR}^2}{\mathrm{~m}}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \text { Work done by gravity } & =m g\left(2 R-R \cos 60^{\circ}\right) \\ & =\frac{3 m g R}{2}\end{aligned}\) \(\begin{aligned} \text { Work done by spring } & =-\frac{1}{2} k\left(0^2-R^2\right) \\ & =\frac{1}{2} k R^2\end{aligned}\) Net work = change in…
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