A position dependent force,\ $F=\left(7-2 x+3 x^{2} ight) \mathrm{N}$ acts on a small body of mass

Question

A position dependent force,\ $F=\left(7-2 x+3 x^{2}ight) \mathrm{N}$ acts on a small body of mass 2 kg and displaces it from $x=0$ to $x=5 \mathrm{~m}$. The work done in joule is\

Professor PC Thomas Classes · Answer

(a) : Force $(F)=7-2 x+3 x^{2}$; Mass $(m)=2 \mathrm{~kg}$ and displacement $(d)=5 \mathrm{~m}$.\ Therefore work done\ $(W)=\int F d x=\int_{0}^{5}\left(7-2 x+3 x^{2}ight) d x=\left[7 x-x^{2}+x^{3}ight]_{0}^{5}$ $=(7 	imes 5)-(5)^{2}+(5)^{3}=35-25+125=135 \mathrm{~J}$\