Question

A 7.25 kg bowling ball is rolled onto a perfectly level surface at a velocity of 10 m/s. The coefficient 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
(A) 1/2km

Answer 1

Hint
Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Complete step by step answer
According to the question given that,
Mass of the bowling ball m = 7.25kg,
The velocity of the bowling ball u = 10m/s ,
Coefficient of friction μ = 0.0025
According to the formula force of the friction,
 $\Rightarrow F = \mu mg $
Putting the values in the equation, and let the value of g = 10 m/s2
 $ \therefore F = 0.0025 \times 7.25 \times 10\,N $
Now, retardation due to force of friction:
 $a = \dfrac{F}{m} = \dfrac{{0.0025 \times 7.25 \times 10}}{{7.25}} = 0.025\,m/{s^2}$
 Let s be the distance travelled by ball before stop.
Using ${v^2} = {u^2} - 2as$
where final velocity of the ball v = 0
∴ we get $ s = \dfrac{{{u^2}}}{{2a}} $
 $ \Rightarrow s = \dfrac{{10 \times 10}}{{2 \times 0.025}} = 2000\,m $
 $ \therefore s = 2\,km $
Thus the ball travels 2 km distance before coming to rest.
Option (C) is correct.

Note
The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.