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=(\mu mgcos\theta )s $
$ =(\mu mgcos\theta )( \frac{h}{\sin \theta } )=\mu mgh\cot \theta $
Now $ \cot {\theta_1}=\cot 30{}^\circ =\sqrt{3} $
And $ \cot {\theta_2}=\cot 60{}^\circ =\frac{1}{\sqrt{3}} $
$ \therefore W_1>W_2 $ i.e., kinetic energy in first case will be less. $ (K=mgh-W) $ Or $ K_1<K_2 $
BETA
