Topic: Engineering Mechanics
A block is made to slide down an inclined plane ($30^{\circ}$ with horizontal) which is smooth. It starts sliding from rest and takes a time 't' to reach the bottom of the plane. An identical body is freely dropped from the same point. The time the body takes to reach the bottom is
- A. t
- B. t/2
- C. t/3
- D. 0
Answer: B. t/2
Explanation: For a block sliding down a smooth incline from height $h$, the distance traveled along the incline is $S = \frac{h}{\sin 30^{\circ}} = 2h$. The acceleration is $a = g \sin 30^{\circ} = \frac{g}{2}$. Using the kinematic equation $S = \frac{1}{2} a t_1^2$, we have $2h = \frac{1}{2} \left(\frac{g}{2}\right) t_1^2$, which gives $t_1 = \sqrt{\frac{8h}{g}} = 2\sqrt{\frac{2h}{g}}$. For a freely falling body dropped from height $h$, $h = \frac{1}{2} g t_2^2$, meaning $t_2 = \sqrt{\frac{2h}{g}}$. Therefore, comparing the two, $t_2 = \frac{t_1}{2}$.