SATHEE: Work Energy And Power Question 267

Work Energy And Power Question 267

Question: A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant $k = 4 m g I R$ , where 2 R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2 R from the base of the rod. If the mass is released at a height 1.5 R, then the velocity of the ring as it reaches the ground is Critical Thinking

Options:

A) $\sqrt{g R}$

B) $2 \sqrt{g R}$

C) $\sqrt{2 g R}$

D) $\sqrt{3 g R}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Initial length of the spring, $ℓ_{i}, = \sqrt{{(2 R)}^{2} + {(1.5 R)}^{2}} = 2.5 R$
$∴ x_{i} = 2.5 R - 2 R = 0.5 R .$

$N o w \frac{1}{2} k x_{i}^{2} + m g (1.5 R) = \frac{1}{2} m v^{2}$ or $\frac{1}{2} k {(0.5 R)}^{2} + m g (1.5 R)$

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Work Energy And Power Question 267

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