A force of 1 N acts on a body of mass 0.5 kg initially at rest. The ratio of the works done by the force in the first, second and third second is :

Options

A. 1:2:3
B. `1:2:5`
C. `1:3:5`
D. `1:5:9`

Answer

c

Explanation

Here applying v=u+at
Also `a=1//0.5=2 ms^(-2)`
velocity after one second
`v_1=0+2xx1=2 ms^(-1)`
Similarly `v_2=0+2xx2=4 ms^(-1)`
`v_3=0+2xx3=6 ms^(-1)`
Change in K.E. after 1 second
`E_1=1/2xx0.5(4-0)=1 J`
Similarly `E_2=1/2xx0.5(16-4)=3J`
`E_3=1/2xx0.5(36-16)=5 J`
Now work done = Change in K.E.
`:.` Required ratio for first three second will be `E_1:E_2:E_3` or `1:3:5`