Question

A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 newtons in OF direction perpendicular to OE. The distance of the body from O after 4 seconds will be

Answer 1

$\underset{–}{AlongOE}$

Velocity, $V_{OE}=3m/s$

Since, applied force is perpendicular to OE so the velocity $V_{OE}$ will be constant.

So, displacement along OE in 4 sec $S_{OE}=3\times 4=12m$

$\underset{–}{AlongOF}$

Force applied, $F=4N$

Mass of the body, $m=2kg$

So, acceleration $a=\frac{F}{m}=2m/s^2$

Displacement along OF in time $t=4sec$

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

$S_{OF}=0+\frac{1}{2}\times 2\times 4^2$

$S_{OF}=16m$

After $t=4sec$, the distance of the body from O to OP

$O{P}^2={12}^2+{16}^2$

$O{P}^2=144+256$

$O{P}^2=400$

$OP=20m$