A force $F=20+10 y$ acts on a particle in $y$-direction where $F$ is in newton and $y$ in meter. Wor

Question

A force $F=20+10 y$ acts on a particle in $y$-direction where $F$ is in newton and $y$ in meter. Work done by this force to move the particle from $y=0$ to $y=1 \mathrm{~m}$ is\

Professor PC Thomas Classes · Answer

(d) : Given : $F=20+10 y$\ work done, $W=\int F . d y$\ $=\int_{0}^{1}(20+10 y) d y=\left[20 y+\frac{10}{2} y^{2}ight]_{0}^{1}=20+\frac{10}{2}=25 \mathrm{~J}$