Question

Find the distance travelled by a body before coming to rest if moving along the horizontal surface with a velocity of $5m{s^{-1}}$. (${\mu _k} = 0.20,{\text{ }}g = 9.8m{s^{ - 2}}$).

Answer 1

Hint: Visualize all the forces on the body and how they affect its motion. The only force hampering the movement of the body is friction. It opposes the relative motion of the body with the surface. In the horizontal direction, the only force acting on the body is the frictional force.

Complete step by step solution:
The frictional force is of two types:
1. Static friction: It is the frictional force acting on the body when it is at rest. In other words, static friction accomplishes to forbid the motion of the body on its surface.
2. Kinetic friction: It is the frictional force acting on the body when it is in motion. It acts in the direction opposite to the motion of the body with respect to its surface.
The kinetic frictional force on the body causing its deceleration is given by:
${f_k} = {\mu _k}N$
where ${\mu _k}$ is the coefficient of kinetic friction
$N$ is the normal force acting on the body
For the object travelling under gravity, the normal force acting on it is equal to the gravitational force acting on it.
$\Rightarrow N = mg$
Thus, the frictional force is:
$\Rightarrow {f_k} = {\mu _k}(mg)$
$\therefore F = m \times a = {\mu _k}(mg)$
Net acceleration of the body,
$\Rightarrow a = \dfrac{{{\mu _k}(mg)}}{m}$
$\Rightarrow a = {\mu _k}g$
Using equations of motion:
$\Rightarrow {v^2} - {u^2} = 2as$
$\Rightarrow 0 - {u^2} = 2( - {\mu _k}g)s$
$\Rightarrow \dfrac{{ - {u^2}}}{{ - 2{\mu _k}g}} = s$
$\Rightarrow s = \dfrac{{{{(5)}^2}}}{{2 \times 0.20 \times 9.8}}$
$\Rightarrow s = \dfrac{{25}}{{3.92}}$
$\Rightarrow s \approx 6.34m$
Here,
$v$ = final velocity of the body = $0$
$u$ = initial velocity f body = $5m{s^{ - 1}}$
$a$ = net acceleration of the body
$s$ = total displacement of the body

The distance travelled by the body before coming to rest is $6.34{\text{ }}m$.

Note: The acceleration is taken with a negative sign because it causes retardation of the body and is actually deceleration. It is really helpful to draw a free body diagram while solving problems related to forces acting on a body.