Question

A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.253. (Indicate the direction with the signs of your answers.)
a.) What is the force of kinetic friction (in N) acting on the block?
(b What is the block's acceleration (in m/s2)?

(c) How far will it slide (in m) before coming to rest?

Answer 1

Explanation:

Given that,

Mass of the block, m = 12.2 kg

Initial velocity of the block, u = 6.65 m/s

The coefficient of kinetic friction, $\mu_k=0.253$

(a)The force of kinetic friction is given by :

$f=\mu_k mg$

mg is the normal force

So,

$f=0.253\times 12.2\times 9.8\\\\f=30.24\ N$

(b) Net force acting on the block in the horizontal direction,

f = ma

a is the acceleration of the block

$a=\dfrac{f}{m}\\\\a=\dfrac{30.24}{12.2}\\\\a=2.47\ m/s^2$

(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(6.65)^2}{2\times 2.47}\\\\d=-8.95\ m$

Hence, this is the required solution.