JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be:                                    [JEE Main 2017]

    A) 9 J

    B) 18 J

    C) 4.5 J

    D) 22 J

    Correct Answer: C

    Solution :

    [c] \[F=6t=ma\]
                \[\Rightarrow \]   \[a=6t\]
    \[\Rightarrow \]   \[\frac{dv}{dt}=6t\]
    \[\int_{0}^{v}{dv}=\int_{0}^{1}{6t}\,dt\]
    \[v=(3{{t}^{2}})_{0}^{1}=3\,m/s\]
    From work energy theorem
    \[{{W}_{F}}=\Delta K.E=\frac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)\]
    \[=\frac{1}{2}(1)(9-0)=4.5\,J\]


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