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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Critical Thinking

  • question_answer
    The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation \[t=\sqrt{x}+3\], where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is           [IIT 1979]

    A)             9 J      

    B)             6 J

    C)             0 J      

    D)             3 J

    Correct Answer: C

    Solution :

                    \[x={{(t-3)}^{2}}\]Þ \[v=\frac{dx}{dt}=2(t-3)\]             at \[t=0\]; \[{{v}_{1}}=-6\,m/s\] and at \[t=6\,\sec \], \[{{v}_{2}}=6\,m/s\]             so, change in kinetic energy\[=W=\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}=0\]


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