Question

The magnitude of the force (in newtons) acting on a body varies with time $t$ (in microseconds) as shown in the figure. $A B, B C$ and $C D$ 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 $=\int F d t=$ area under $F-t$ graph
$\therefore$ Total impulse from $t=4 \mu$ s to $t=16 \mu s$
$=$ Area $E B C D$
$=$ Area of trapezium $E B C F+$ Area of triangle $F C D$
$=\frac{1}{2}(200+800) 2 \times 10^{-6}+\frac{1}{2} \times 800 \times 10 \times 10^{-6} $
$=5 \times 10^{-3} \,N - s$