A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. If it starts its journey from rest at `x=0`, its velocity at `x=12m` is

Options

A. `0(m)//(s)`
B. `20sqrt2(m)//(s)`
C. `20sqrt3(m)//(s)`
D. `40(m)//(s)`

Answer

D

Explanation

`x=0` to `x=12m`
`W=`area of F-x graph
`=(1)/(2)(4+12)xx10=80J`
`x=0`,`K_1=0`
`W=K_2-K_1`
`80=(1)/(2)mv^2-0=(1)/(2)xx0.1v^2`
`v^2=1600impliesv=40(m)//(s)`