Question

A cricket ball is rolled on ice with a velocity of $5.6m/s$ and comes to rest after traveling $8m$. Find the coefficient of friction. Given $g=9.8m/s^2$

Answer 1

Since the ball comes to rest in 8m, from the second equation of motion,

$v^2=u^2+2\times a\times s$

Put the value of final velocity as 0 and the distance as $8m$.

We get the value of the required acceleration to do this job successfully as a $=-1.96\frac{m}{s^2}$

(The negative sign signifies deceleration)

Now, we know that the force of friction is given by $\mu N$, where N is the normal reaction force exerted by the surface on the ball, and μ the coefficient of friction of the surface.

So, the acceleration produced by frictional force = $\frac{\mu N}{M}$, where $M$ is the mass of the ball.

But, $N=Mg$. where $g$ is the gravitational acceleration,

Thus, the acceleration produced$=\mu g$. Mass cancels out.

Thus, from the first calculation,$\mu g=1.96$

So,$\mu =\frac{1.96}{9.8}=0.2$