Question

A force is applied on a body of mass 0.9kg that is at rest the .The force is applied for a duration of 5seconds and as a result the body covers a distance of 250m .find the magnitude of the force

Answer 1

Distance covered (s)= 250m
Time (t) = 5s
Inital velocity (u) = 0


Using 2nd equation of motion let's find the acceleration first

s= ut + (1/2)at^2
250= 0 + (1/2) a (5)^2
250= (1/2) a (25)
(250 )(2)/(25) = a
20 =a

So acceleration is 20m/s^2


Now,
Mass of that body = 9/10 kg.
Acceleration = 20m/s^2

Force = mass × acceleration
= (9/10)(20)
= 18N.

Answer 2

Answer:

The magnitude of the force applied on a body is equal to 18N.

Explanation:

Given: the initial velocity of body, $u=0$   (body at rest)

The time for which force applied, $t=5sec$

The body covers distance, $S=250m$

Use 2nd equation of motion to find acceleration:

$S=ut+\frac{1}{2}at^{2}\\ 250=0+ \frac{1}{2}a(5)^{2}\\ 250=\frac{25a}{2} \\ a=20ms^{-2}$

Now given the mass of the body, $m=0.9kg$

Therefore the magnitude of force:

$F=ma\\ F=(0.9kg)(20ms^{-2})\\ F=18kgms^{-2}\\ F=18N$

Hence the force applied on the body would be 18N.