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 0.5. The distance that the box will move relative to belt before coming to rest on it (taking $g = 10m{s^{ - 2}}$), is:
(a). 1.2m
(b). 0.6m
(c). Zero
(d). 0.4m

Answer 1

Hint: The applied force is equal to the product of the mass of the body and the acceleration produced in the body. The force here is due to the friction between body and the horizontal plane and for displacement, we can apply equations of motion

Complete step by step answer:
Frictional force is a force which comes into action when two bodies slide over one another. This force is large when the surfaces are rough and small for smooth surfaces. The amount of friction that a surface exhibits can be characterized by what we call a coefficient of friction.
Coefficient of friction is defined as the ratio of the frictional force between two surfaces and the force that is keeping in contact with each other. The expression for frictional force is
$F = \mu mg$
where $\mu$ is the coefficient of friction and it is dimensionless and does not have any units.
We are given
U = 2m/s
$\mu$= 0.5
$g = 10m{s^{ - 2}}$

Force in the presence of friction is given as
$F = \mu mg$

Also, F = ma
$\therefore ma = \mu mg \\\ \Rightarrow a = \mu g \\\ \Rightarrow a = 0.5 \times 10 \\\ \Rightarrow a = 5m{s^{ - 2}} \\\$
As we know,
${{\text{v}}^2} - {u^2} = 2aS$

Since the box finally comes to rest, v = 0 and we need to find distance S. Also, the bag is decelerating so a is negative.
$0 - {u^2} = 2aS \\\ \Rightarrow - 4 = - 2 \times 5 \times S \\\ \Rightarrow S = \dfrac{4}{{10}} = 0.4m \\\$

Hence, the distance travelled by bag is equal to 0.4m before it comes to rest and option d is correct.

Note: In practical applications, friction is always there between two bodies and $F = \mu mg$ is used as a general expression. F = mg is considered a special case when there is no friction between bodies and coefficient of friction is equal to 1.