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NEET Sample Paper NEET Sample Test Paper-53

  • question_answer
    The distance x moved by a body of mass 0.5 kg by a force varies with time t as \[x=3{{t}^{2}}+4t+5\] where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

    A) 25 J                             

    B) 50 J

    C) 75 J                             

    D) 100 J

    Correct Answer: C

    Solution :

    \[Velocity\text{ }\left( v \right)=\frac{dx}{dt}=\frac{d}{dt}\left( 3{{t}^{2}}+4t+5 \right)=6t+4\] Acceleration is a \[a=\frac{dv}{dt}=\frac{d}{dt}\left( 6t+4 \right)=6\,m{{s}^{-2}}\] Therefore, applied force is \[F=ma=0.5\times 6=3N\] Now\[t\,\,=\,\,2s\], the distance moved is \[x=3\times {{\left( 2 \right)}^{2}}+4\times 2+5=25\,m\] \[\therefore \,\,\,Work\text{ }done\text{ }W=Fx=3\times 25=75\,J.\] Hence the correct choice is [c].


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