Question

A $7.25$ kg bowling ball is rolled onto a perfectly level surface at a velocity of $10m/s$. The co-efficient of friction between the surface and the bowling ball is $0.0025$. If the surface is perfectly level and is long enough, how far will the bowling ball roll before it comes to a complete stop?

• A. $20m$
• B. $200m$
• C. $2km$
• D. $20km$
• E. $1/2km$

Answer 1

The correct option is C $2km$

Given : $m=7.25kg,u=10m/s,\mu =0.0025$

Force of friction, $F=\mu mg$

$\therefore$ $F=0.0025\times 7.25\times 10N$

Now, retardation due to force of friction:

$a=F/m=0.0025\times 7.25\times 10/7.25=0.025m/s^2$

Let $s$ be the distance travelled by ball before stop.

Using $v^2=u^2-2as$

where final velocity of the ball $v=0$

$\therefore$ we get $s=\frac{u^2}{2a}$

$s=\frac{10\times 10}{2\times 0.025}=2000m$

$s=2km$

Thus the ball travels $2km$ distance before coming to rest.