Question

A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?

Answer 1

Friction force acting on the block will decelerate it.
Let the deceleration be 'a'.

Using free body diagram


R − mg = 0
(where R is the normal reaction force)
⇒ R = mg (1)
Again, ma − μ_kR = 0
(where μ_k is the coefficient of kinetic friction)
From Equation (1),
⇒ ma = μ_kmg
⇒ a = μ_kg = 0.1 × 10
= 1 m/s²
Given:
initial velocity, u = 10 m/s
final velocity, v = 0 m/s (block comes to rest)
a = −1 m/s² (deceleration)
Using equation of motion v² $-$ u² = 2as
(where s is the distance travelled before coming to rest)
$s=\frac{v^2-u^2}{2a}$
On substituting the respective values, we get
$$\begin{gathered} =0-\frac{{10}^2}{2}\left(-1\right) \\ =\frac{100}{2}=50\mathrm{m} \end{gathered}$$
Therefore, the block will travel 50 m before coming to rest.