Question

A block of mass 2.9 kg slides over a frictional floor. If it slides 10m in 5 s before coming to rest. Calculate the force of friction on the ball.

• A. - 4N
• B. - 1.6N
• C. - 0.8N
• D. - 5N
• E. - 2.32N

Answer 1

Answer: (E)

- 2.32N

Given : $m=2.9kg$ $S=10$ m $t=5$ s $v=0$ m/s

Let the magnitude of friction force be $f$ and initial velocity of the ball be $u$.

Thus the acceleration of the ball $a=\frac{-f}{m}$

Using $v=u+at$

$\therefore$ $0=u-\frac{f}{m}t$ $⟹u=\frac{ft}{m}=\frac{f\times 5}{2.9}=1.72f$

Also $S=ut+\frac{1}{2}a{t}^2$

$\therefore$ $10=\left(1.72f\right)\times 5-\frac{1}{2}\times \frac{f}{2.9}\times 5^2$

OR $10=8.6f-4.3f$ $⟹f=2.32$ N

As the frictional force acts in the opposite direction, thus frictional force is $-2.32$ N