• # question_answer A position dependent force $F=7-2x+3{{x}^{2}}\,newton$ acts on a small body of mass 2 kg and displaces it from $x=0$ to $x=5\,m$. The work done in joules is         [CBSE PMT 1994] A)             70        B)             270 C)             35        D)             135

$W=\int\limits_{0}^{5}{Fdx}=\int\limits_{0}^{5}{(7-2x+3{{x}^{2}})\ dx}$ = ${{[7x-{{x}^{2}}+{{x}^{3}}]}^{5}}$ = 35 ? 25 + 125 = 135 J