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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Work Done by Variable Force

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    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

    Correct Answer: D

    Solution :

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


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