Question

A body is moving with a velocity of 36 kmph on a horizontal surface having coefficients of frictional 0.2
the distance travelled by it before it comes to rest is

Answer 1

Answer:

u = 36*5/18 = 10m/s

v=0

ma = coefficient *mg (Motion provided by friction)

a = coefficient* g

a = 0.2*10 = 2m/s square

s = u squared / 2a

= 10*10/2*2 = 100/4= 25m

Answer 2

The distance travelled by the body is 25 meters.

Explanation:

Given that,

Initial speed of the body, u = 36 km/h = 10 m/s

It comes to rest, v = 0

The coefficient of friction, $\mu=0.2$

In this case,

$ma=-\mu mg$

$a=\mu g$

$a=-0.2\times 10\ m/s^2=-2\ m/s^2$

Let d is the distance travelled by it before it comes to rest. It is given by using third equation of motion as :

$v^2-u^2=2ad$

$-u^2=2ad$

$d=\dfrac{-u^2}{2a}$

$d=\dfrac{-(10)^2}{2\times (-2)}$

d = 25 meters

So, the distance travelled by the body is 25 meters. Hence, this is the required solution.

Learn more,

Equation of motion

https://brainly.in/question/683267