JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • 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 Online Paper (Held On 09-Jan-2019 Evening]

    A) 950 J                                       

    B) 900 J

    C) 875 J               

    D)               850 J

    Correct Answer: B

    Solution :

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


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