Laws Of Motion Question 325
Question: A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is$ \mu =0.5 $ . The distance that the box will move relative to belt before coming to rest on it taking $ g=10m{{s}^{-2}} $ , is
Options:
A) 1.2 m
B) 0.6 m
C) zero
D) 0.4 m
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Frictional force on the box $f=\text{ }\mu mg$
$\therefore \text{ Acceleration in the box a = }\mu \text{g =5m}{{\text{s}}^{-2}}$
${{v}^{2}}={{u}^{2}}+2aS\Rightarrow 0={{2}^{2}}+2\times \left( 5 \right)s\Rightarrow s=-\frac{2}{5}$
$\text{w}\text{.r}\text{.t}\text{. belt }\Rightarrow \text{ distance = 0}\text{.4m}$
