Question

A bowling ball with an initial velocity of 3 m/sec rolls along a level floor for
50m before coming to a stop. What is the coefficient of rolling friction?

Answer 1

Given:

A bowling ball with an initial velocity of 3 m/sec rolls along a level floor for 50m before coming to a stop.

To find:

Coefficient of rolling friction ?

Calculation:

• Frictional acceleration will be $\mu g$, here $\mu$ is the coefficient of rolling friction.

Applying EQUATIONS OF KINEMATICS:

${v}^{2} = {u}^{2} + 2as$

$\implies \: {0}^{2} = {3}^{2} + 2( - \mu g)50$

$\implies \: {0}^{2} = {3}^{2} - 2( \mu g)50$

$\implies \: 100 \mu g = 9$

$\implies \: 1000 \mu = 9$

$\implies \: \mu = 0.009$

So, coefficient of rolling friction is 0.009.

Answer 2

$\underline{ \: \: \bold{\orange {Solution :-}}}$

By Using Equation of Kinematics,

v² - u² = 2as

⇒ (0)² = (3)² - 2(-μg)50

⇒ 100μg = 9

⇒ 1000μ = 9

⇒ μ = 0.009

So, The co-efficient of the Rolling Friction is 0.009