The velocity of a block of mass `2kg` moving along `x-` axis at any time `t` is givne by `v=20-10t(m//s)` where `t` is in seconds and `v` is in `m//s`. At time `t=0`, the block is moving in positive `x-` direction.
The work done by net force on the block starting from `t=0` till it covers a distance 25 metres will be `:`

Options

A. `+200`
B. `-200J`
C. `+300J`
D. `-300J`

Answer

D

Explanation

The velocity of particle is zero when `v=(20-10t)=0`.
That it at `t=2 sec. v=0`

From `t=0` to `t=2` distance travel
`S_(1)=((20)^(2))/(2xx10)=20m.`
Next 5 meter will be covered in `5=(1)/(2)xx10xxt^(2) or t=1s`.
`:. ` The particle covers 25 meters distance in 3 sec.
`K.E. at t=0` is `K_(i)=(1)/(2)m u^(2)=(1)/(2) 2xx(20)^(2)=400J`
KE at `t=3` is
`K_(f)=(1)/(2)mv^(2)=(1)/(2)2xx(10)^(2)=100J`
Therefore work done by block from `t=0` to `t=3s ` is
`DeltaW=K_(f)-K_(i)=100-400=-300J`