Question

A force of `4N` acts on a body of mass `2kg` for `4s`. Assuming the body to be initially at rest, find (a) its velocity when the force stops acting (b) the distance covered in `10s` after the force starts acting.

Answer 1

(a)Here, `F=4N, m=2kg, t=4s, u=0, v=?`
From `F=ma, a=F/m= 4/2= 2m//s^(2)`
From `v=u+at, v=0+2xx4= 8m//s`
(b) Distance travelled in first `4s` when the force is actually acting, `s_(1)= ut+1/2 at^(2)= 0+1/2xx2xx4^(2)= 16m`
When the force stops acting, velocity of the body is constant, `v=8m//s`
Distance travelled in the last 6 second,`s_(2)=vxxt= 8xx6=48m`
Distance covered in `10 s` after the force starts acting `=s_(1)+s_(2)= 16+48= 64` m.

Comments

S2=      v*t 8*6 t time how come 6

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After the distance traveled in 4s, time of 6s  is left(from total of 10s) in which we have to find the distance