Question

The magnitude of the force (in newtons) acting on a body varies with time $t$ (in microseconds) as shown in the figure. $AB,BC$ and $CD$ are straight line segments. The magnitude of the total impulse of the force on the body from $t=4\mu s$ to $t=16$ us is ......... $N-s.$

Answer 1

Impulse $=\int Fdt=$ area under $F-t$ graph
$\therefore$ Total impulse from $t=4\mu$ s to $t=16\mu s$
$=$ Area $EBCD$
$=$ Area of trapezium $EBCF+$ Area of triangle $FCD$
$=\frac{1}{2}(200+800)2\times 10^{-6}+\frac{1}{2}\times 800\times 10\times 10^{-6}$
$=5\times 10^{-3}N-s$