Question

(1)
Q10. A ball rolled on ice formed on the surface of a frozen lake with an initial speed of 1.8 m/s
goes on moving for a total distance of 12m before stopping. Find the value of the
coefficient of rolling friction for ball-ice pair

(1)​

Answer 1

Explanation:

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

v

2

=u

2

+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 =−1.96

s

2

m

(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 =

M

μN

, 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=

9.8

1.96

=0.1