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

  • question_answer
    A force acts on a 2 kg object so that its position is given as a function of time as\[x=3{{t}^{2}}+5\]. What is the work done by this force in first 5 seconds?   [JEE Main 09-Jan-2019 Evening]  

    A) 950 J

    B) 900 J

    C) 875 J

    D) 850 J

    Correct Answer: B

    Solution :

    [b] \[x=3{{t}^{2}}+5\]
    \[V=\frac{dx}{dt}=6t\]
    at \[t=0,\,\,u=0\]
    at \[t=5,\,\,v=30m/s\]
    \[W=\Delta \,K=\frac{1}{2}\,\,\times \,\,2{{\left( 30 \right)}^{2}}=900\text{ }J\]
    Option [b] is correct.


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