SATHEE: Work Energy And Power Question 141

Work Energy And Power Question 141

Question: A heavy particle hanging from a string of length $l$ is projected horizontally with speed $\sqrt{2 g l}$ . The speed of the particle at the point where the tension in the string equals weight of the particle

Options:

A) $\sqrt{2 g l}$

B) $\sqrt{3 g l}$

C) $\sqrt{g l / 2}$

D) $\sqrt{g l / 3}$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $T - m g \cos θ = \frac{m v^{2}}{R}$ Given $T = m g$

$m g - m g \cos θ = \frac{m v^{2}}{R}$

$g (1 - c o s θ) = \frac{v^{2}}{R}$ C.O.M.E. at A and B; $Δ K + Δ U = 0$

$(\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}) + m g (R - R c o s θ) = 0$

$\Rightarrow v^{2} - u^{2} = - 2 g R (1 - c o s θ)$

$\Rightarrow v^{2} - {(\sqrt{g l})}^{2} = - 2 v^{2}$

$3 v^{2} = g l \Rightarrow v = \sqrt{\frac{g l}{3}}$

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Work Energy And Power Question 141

Question: A heavy particle hanging from a string of length l is projected horizontally with speed 2gl . The speed of the particle at the point where the tension in the string equals weight of the particle

Options:

Answer:

Solution:

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Question: A heavy particle hanging from a string of length $l$ is projected horizontally with speed $\sqrt{2 g l}$ . The speed of the particle at the point where the tension in the string equals weight of the particle