Question

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

• A. 135
• B. 270
• C. 35
• D. 70

Answer 1

Answer: (A)

Force $(F)=7-2 x+3 x^{2} ; $
Mass $(m)=2 kg$ and displacement $(d)=5 m$.
Therefore work done
$(W)=\int F d x=\int\limits_{0}^{5}\left(7-2 x+3 x^{2}\right) d x=\left(7 x-x^{2}+x^{3}\right)_{0}^{5} $
$=(7 \times 5)-(5)^{2}+(5)^{3}=35-25+125=135\, J$