AMU Medical AMU Solved Paper-2013

  • question_answer
    Figure gives the acceleration of a 2.0 kg body as it moves from rest along x axis while a variable force acts on it from \[x=0\,m\] to\[x=9\,m\]. The work done by the force on the body when it reaches (i) \[x=4\,m\] and (ii) \[x=7\,m\] shall be as given below

    A)  21 J and 33 J respectively

    B)  21 J and 15 J respectively

    C)                  42 J and 60 J respectively

    D)  42 J and 30 J respectively

    Correct Answer: D

    Solution :

                     From figure Work done (W) = max                 \[{{W}_{4}}=2\left[ \frac{1}{2}\times 1\times 6+3\times 6 \right]\]           = 42 J                 \[{{W}_{7}}=2\,[21+3-3-1\times 6]\]\[=30\,J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner