Question

A block slides on a horizontal surface with an initial speed of 10 m/s. If the block takes 2s to come to a stop, what is the coefficient of kinetic friction between the block and the surface? (The answer does not depend on the mass of the block.) (A) 0.1 (B) 0.2 (C) 0.4 (D) 0.5

Answer 1

Answer:

The coefficient of kinetic friction between the block and the surface is 0.5

Explanation:

It is given that,

Initial velocity of the block, u = 10 m/s

Time taken by the block to come to rest, t = 2 s

So, final velocity, v = 0

We need to find the coefficient of kinetic friction between the block and the surface. According to second law of motion :

F = ma

And friction force F = -μmg

i.e.

$\mu mg=ma$

$\mu=\dfrac{a}{g}$...........(1)

Firstly, we will find the value of a i.e. acceleration

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-10\ m/s}{2\ s}$

a = -5 m/s²

So, equation (1) becomes :

$\mu=\dfrac{5\ m/s^2}{9.8\ m/s^2}$

$\mu=0.5$

So, the coefficient of kinetic friction between the block and the surface is 0.5. hence, this is the required solution.