Question

A body of mass $0.40kg$ moving initially with a constant speed of $10m{s}^{-1}$ to the north is subject to a constant force of $8.0N$ directed towards the south for $30s$. Take the instant the force is applied to be $t=0$, the position of the body at that time to be $x=0$, and predict its position at $t=5s,25s,100s$.

Answer 1

Given:

The mass of body, $m=0.40kg$

Constant initial speed of the body is, $u=10m/s$

Force that acts on the body is, $F=8.0N$

$\left(i\right)$

At $t=–5s$

The force starts acting on the body from $t=0s$.

So, the acceleration of the body during this time was zero and it moves at a constant speed.

Position of the body is given by:

$x=v\times t$

$x=10\times (-5)$

$=-50m$

$\left(ii\right)$

At $t=25s$

Since the force acts in the opposite direction of motion of particle. So, the acceleration of the body due to the force acting on it is:

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

$=\frac{-8.0}{0.40}={-20ms}^{-2}$

The position is given by the second equation of motion:

$s^{\prime }=u{t}^{\prime }+\frac{1}{2}a"$$t^2$

$=10\times 25+\frac{1}{2}\times (-20)\times {25}^2$

$=250-6250=-6000m$

$\left(iii\right)$

At $t=100s$

For first 30 sec, it moves under retardation of force and after that the body moves at constant speed.

$a=-20m$$s^{-2}$

$u=10m/s$

The distance covered in $30s$:

$s_1=ut+\frac{1}{2}a"t^2$

$=10\times 30+\frac{1}{2}\times (-20)\times {30}^2$

$=300-9000=-8700m$

Now, the speed after $30sec$ is:

$v=u+at$

$v=10-(20\times 30)$

$=-590m/s$

Distance covered in remaining 70 sec:

$x^{\prime }=-590\times 70$

$=-41300m$

So, total distance covered in $100s$ will be:

$X=-8700-41300$

$X=-50000m=-50km$