Question

A constant force act on an object of the mass of 7 kg for a duration of 3 second it increased the objects velocity from 5 metre per second to 6 metre per second find the magnitude of the applied force

Answer 1

Answer :-

2.31N

Explanation :-

Given :

• Mass of the object,m = 7kg

• Time taken,t = 3 seconds

• Initial velocity,u = 5m/s

• Final velocity,v = 6m/s

To Find :

• Force,F =?

Solution :

We know,

$\boxed{\sf{}F=ma}$

We do know the acceleration,so let’s find out it first.

According to the first equation of motion,

$\boxed{\sf{}v=u+at}$

Put their values and solve it,

$\implies \sf{}6m/s=5m/s+a\times 3s$

$\implies \sf{}6m/s-5m/s=a\times 3s$

$\implies \sf{}1m/s=a\times 3s$

$\sf{}\implies a=\dfrac{1m/s}{3s}$

$\sf{}\implies a=0.33m/s^2$

Therefore,acceleration is equal to 0.33m/s^2

Force,

$\sf{}\implies Mass\times Acceleration$

$\sf{}\implies 7kg\times 0.33m/s^2$

$\sf{}\implies 2.31N$

Hence,magnitude of applied force is equal to 2.31N