Question

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the mass of the body?

Answer 1

From the question,

`u = 0m {s}^( - 1)`

`v = 6m {s}^( - 1)`


`t = 3s`

`F=12N`



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.


`Ft =m(v-u)`



`12(3.0)=m(6.0- \: 0)`
This implies that


`36.0 = 6m`

`m = (36.0)/(6.0)`

`\therefore \: m = 6.0kg`


You can also use the equation of linear motion,

`v = u + at`

`6 = 0 + a(3)`

`6 = 3a`

`a = (6)/(3)`


`a = 2 {ms}^( - 2)`
But

`F=ma`

`12 = m(2)`

`12 = 2m`

`(12)/(2) = m`

`\therefore \: m = 6kg`