Question

A particle of mass $0.2kg$ is kept at rest. A force of $2N$ acts on it for $2$ seconds. Find the distance moved by the particle in $\left(i\right)$ these $2$ seconds and $\left(ii\right)$ next $2$ second if the same force is continued to be applied.

• A. $20m;20m$
• B. $20m;60m$
• C. $20m;40m$
• D. $40m;20m$

Answer 1

Answer: (B)

$20m;60m$

Given,

Mass of the particle, $m=0.2kg$

Force acting, $F=2N$

Initial velocity, $u=0m/s$

From 2nd law of newton's,

$F=ma$

$a=\frac{F}{m}$

$a=\frac{2}{0.2}=10m/s^2$

(1) Time, $t=2sec$

From 2nd equation of motion,

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

$S=0\times 2+\frac{1}{2}\times 10\times 2\times 2$

$S=20m$

(ii) For next $2sec$,

The final velocity will be,

$v=u+at$

$v=0+10\times 2=20m/s$

Hence the initial velocity of the particle for next $2sec$ is $20m/s$.

From the 2nd equation of motion,

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

$S=20\times 2+\frac{1}{2}\times 10\times 4$

$S=40+20=60m$

The correct option is B.