The magnitude of force (in newtons) acting on a body varies with timer `t` (in microseconds) as shown in Fig. `AB, BC` and `CD` are straight line segments. The magnitude of total impulse of the force on the body from `t=4s` to `t=16 s` is____________`Ns`

Answer

0.005

Explanation

We know that area under the F-t graph gives the impulse imparted to the body. The magnitude of total impulse of force on the body from
`t = 4 mu s "to" t = 16 mu s" = "area" (BCDFEB)`
= area of BCFEB + area CDFC
`1/2 (200 + 800) xx 2 xx 10^(-6) + 1/2 xx 10 xx 800 xx 10^(-6)`
`=0.001 + 0.004 = 0.005 Ns`.