JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Sample Paper Topic Test - Work Energy Power

  • question_answer
    A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from \[{{v}_{1}}\], to \[{{v}_{2}}\]is (neglect friction)

    A) \[\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})\]

    B) \[\frac{m}{3P}({{v}_{2}}-{{v}_{1}})\]

    C) \[\frac{3p}{m}(v_{2}^{2}-v_{1}^{2})\]

    D) \[\frac{m}{3P}(v_{2}^{2}-v_{1}^{2})\]

    Correct Answer: A

    Solution :

    [a] \[P=Fv=mav\]
      \[\Rightarrow \] \[a=\frac{P}{mv}\]
      \[\Rightarrow \] \[v\frac{dv}{ds}=\frac{P}{mv}\]
      \[\Rightarrow \] \[{{v}^{2}}dv=\frac{P}{m}ds\]
      \[\Rightarrow \] \[\frac{P}{m}\int_{\,0}^{\,s}{ds}=\int_{\,{{v}_{1}}}^{\,{{v}_{2}}}{{{v}^{2}}dv}\]
      \[\Rightarrow \] \[\frac{P}{m}s=\frac{1}{3}(v_{2}^{3}-v_{1}^{3})\]
      \[\Rightarrow \] \[s=\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})\]

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