Question

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s, is stopped by friction in 10 s. Then the coefficient of friction is (g = 10 m/s2)
(1) 0.04 (2) 0.06 (3) 0.02 (4) 0.03

Answer 1

Mass of the block = 2 kg

Initial velocity (u) = 6 m/s

Final velocity = 0 m/s

Let the acceleration provided by the friction be a.

Time taken (t) = 10

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

$a = -\frac{6}{10}$

a = - 0.6 m/s^2

Negative means that it is opposite to velocity.

Now,

μg = 0.6 m/s^2

μ = $\frac{0.6}{10}$

μ = 0.06

Hence, the coefficient of friction is 0.06.

Hence, option (2) is correct.