Question:
A force of $\mathrm{F}=(5 \mathrm{y}+20) \hat{\mathrm{j}} \mathrm{N}$ acts on a particle. The workdone by this force when the particle is moved from $\mathrm{y}=0 \mathrm{~m}$ to $\mathrm{y}=10 \mathrm{~m}$ is
Solution:
$\mathrm{F}=(5 \mathrm{y}+20) \hat{\mathrm{j}}$
$\omega=\int F d y=\int_{0}^{10}(5 y+20) d y$
$=\left(\frac{5 y^{2}}{2}+20 y\right)_{0}^{10}$
$=\frac{5}{2} \times 100+20 \times 10$
$=250+200=450 \mathrm{~J}$