Question

A ball of mass $1kg$ is rolled on a ground with a velocity of $20m/s$ and it comes to rest after travelling a distance of $50m$. What is the magnitude of force of friction between the ball and the ground?

• A. $5N$
• B. $3N$
• C. $0N$
• D. $4N$

Answer 1

The correct option is D $4N$

Initial velocity of the ball $u=20m/s$
The final velocity of the ball, $v=0$
(as the stone finally comes to rest)
Distance covered by the stone = $50m$

Using equation of motion:
$v^2=u^2+2as$
$0\times 0=20\times 20+2\times a\times 50$
$a=-4m{s}^{-2}$
Negative value of acceleration means that the speed is decreasing.

Mass of the ball, $m=1kg$

From newton's second law of motion:
$F=ma$
$F=1\times (-4)=-4N$
So, the force of friction between the ball and the ground is -4 N. Here, the negative sign indicates that force is opposing the motion in original direction.

Hence, magnitude of force is $4N$.