Question

A cricket ball leaves the player's bat with a velocity of 10 " and slides 40 m before coming to rest, Calculate coefficient of friction between the ball and the ground?​

Answer 1

Answer:

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

v2=u2+2×a×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 =−1.96s2m

(The negative sign signifies deceleration)

Now, we know that the force of friction is given by μN, 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 = MμN, where M is the mass of the ball.

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

Thus, the acceleration produced=μg. Mass cancels out.

Thus, from the first calculation,μg=1.96

So,μ=9.81.96=0.1