An observer and a vehicle, both start moving together from rest (towards the right) with acceleration `5 ms^(-2)`and `2 ms^(-2)`, respectively. There is a 2 kg block on the floor of the vehicle and coefficient of friction is `mu = 0.3` between their surface. Then the work done by the frictional force on the 2 kg block observed by the running observer, during the first 2 seconds of the motion

Options

A. `24J`
B. `-24J`
C. `16J`
D. `36J`

Answer

B

Explanation

`f_("limiting")=mumg=0.3xx2.10=6N`
Force needed for accn of `2m//s^(2)`
`F=ma=4N`
So friction force is static `f=4N`
displacement w.r.t observer in `t=2` sec is
`S=U_("rel")t+1/2a_("rel") t^(2)=01/2(-3)xx4=-6m`
`W=vecf.vecS=4(-6)=-24J`