An external force of 6N is applied on body at rest at time t=0, at t=1 sec direction of force is reversed. Find the speed of body when it comes back to the initial position

Options

A. zero
B. `sqrt(2) m//s`
C. `sqrt(1.2) m//s`
D. `0.5 m//s`

Answer

C

Explanation

From A to B
`V_(B)=(6-4)/2xx1= 1m//s`
dist. Travelled `S_(AB) ,S_(AB)=1/2xx1xx(1)^(2)=1/2 =0.5 m` from B to C
`0=(1)^(2)-2xx10/2xxS S_(BC)=0.1 m`
`S_(AC)=S_(AB)+S_(BC)=0.6 m`
Velocity when it comes black to A
`V_(A)^(2)=O+2xx(6-4)/2xx0.6`
`V_(A)=sqrt(1.2) m//s`