Question

A body starts sliding on rough horizental surface with speed 10m/s if the coeffcient of friction is 0.2 the distance covered by body before coming to rest is

Answer 1

Answer:25m

Explanation:

I assumed g=10 Nkg-1

F=ma

Friction=μR     R is normal reaction

R=mg

therefore -μmg= ma    because friction acts onto the opposite direction of the motion

                  a= 0.2*10 = 2 ms-2

     v2 =u2 +2as

   0 = 10*10 - 2 *2*s

 s = 25m

Answer 2

Answer:

A body starts sliding on a rough horizontal surface with a speed of 10m/s if the coefficient of friction is 0.2 the distance covered by the body before coming to rest is $25.5m$

Explanation:

When a body travels along a rough surface it moves a distance before it comes to rest, this distance is called stopping distance

for a body moving with velocity $v$ on a rough surface with a coefficient of friction μ the stopping distance is given by the formula

$d=\frac{v^2}{2\mu g}$

where the acceleration due to gravity $g=9.8m/s^{2}$

from the question, it is given that

the velocity of the body $v=10m/s$

the coefficient of friction μ$=0.2$

substitute the given values to get the distance

$d=\frac{10^{2} }{2*0.2*9.8}=25.5m$