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AMU Medical AMU Solved Paper-2013

  • question_answer
    A horizontal cable accelerates a package across a frictionless horizontal floor. The amount of work that has been done by the cables force on the package is given by \[W(t)=(0.20\,J/{{s}^{2}})\,{{t}^{2}}\]. The average power < P > due to cables force in the time interval \[t=5\,s\] to \[{{t}_{2}}=10\,s\] and the instantaneous power at t = 3 s are

    A)  2.0 W, 1.80 W   

    B)  2.0 W, 1.20W

    C)  3.0 W, 1.80W    

    D)  3.0W, 1.20W

    Correct Answer: B

    Solution :

                    We know that                 \[P=\frac{dW}{dt}\]                 \[P=(0.20\times 2)t\]                 \[P=0.4\,t\]                 \[P=0.4\times 5=2.0\,W\]                 \[{{P}_{ins}}=0.4\times 3\]                                 \[=1.2\,W\]


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