SATHEE: Work Energy And Power Question 150

Work Energy And Power Question 150

Question: A particle is released from the top of two inclined rough surfaces of height ’ $h$ ’ each. The angle of inclination of the two planes are $30^{\circ}$ and $60^{\circ}$ respectively. All other factors (e.g. coefficient of friction, mass of block etc.) are same in both the cases. Let $K_{1}$ and $K_{2}$ be the kinetic energies of the particle at the bottom of the plane in two cases. Then

Options:

A) $K_{1} = K_{2}$

B) $K_{1} > K_{2}$

C) $K_{1} < K_{2}$

D) data insufficient

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Work done in friction is $W = (μ m g c o s θ) s$

$= (μ m g c o s θ) (\frac{h}{\sin θ}) = μ m g h \cot θ$

Now $\cot θ_{1} = \cot 30^{\circ} = \sqrt{3}$

And $\cot θ_{2} = \cot 60^{\circ} = \frac{1}{\sqrt{3}}$

$∴ W_{1} > W_{2}$ i.e., kinetic energy in first case will be less. $(K = m g h - W)$ Or $K_{1} < K_{2}$

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Work Energy And Power Question 150

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