Question

A 20kg body is pushed with 7N for 1.5sec then with a force of 5N for 1.7dec and finally with a force of 10N for 3sec,the total impulse applied to the body and change in velocity will be

Answer 1

The impulse is given as:
I = Ft
$$\begin{gathered} I_1=F_1{t}_1=7\times 1.5=10.5\mathrm{N}\mathrm{s} \\ {I}_2=F_2{t}_2=5\times 1.7=8.5\mathrm{N}\mathrm{s} \\ {I}_3=F_3{t}_3=10\times 3=30\mathrm{N}\mathrm{s} \\ \mathrm{Therefore},\mathrm{the}\mathrm{total}\mathrm{impulse}I=I_1+I_2+I_3=10.5+8.5+30=49\mathrm{N}\mathrm{s} \\ \mathrm{Impulse},I=m(v-u) \\ \Rightarrow 49=20(v-0) \\ \Rightarrow v=\frac{49}{20}=2.45\mathrm{m}/\mathrm{s} \end{gathered}$$