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9th Class Science Work and energy Question Bank Work Power and Energy IIT JEE Objective Problems

  • question_answer
    A body of mass 2 kg is thrown up vertically with a K.E. of 490 J. If the acceleration due to gravity is \[\text{9}.\text{8 m}{{\text{s}}^{-\text{2}}}\], the height at which the K.E. of the body becomes half of the original value is

    A)  50 m                                    

    B)  25 m

    C)  12.5 m    

    D)  10 m

    Correct Answer: C

    Solution :

      Given, total energy \[(T.E.)=490\,\,J\] Mass of the body \[(m)=2\,\,kg\] Acceleration due to gravity \[(g)=9.8\,\,m/{{s}^{2}}\] \[h=?\] Given that certain height K.E becomes half of the total energy i.e. \[\frac{490}{2}=245J\] We know that, Total energy (TE) = kinetic energy (K.E) + potential energy (P.E) \[\Rightarrow 490=245+P.E.\] \[\Rightarrow P.E.=490-245=245\,J\] We also now that\[P.E.=mgh\] Now we have to find ?h? ? Applying the the above values in equation (1) we get, \[\Rightarrow 245=2\times 9.8\times h\] \[\Rightarrow h=\frac{245}{2\times 9.8}=12.5m\]                            


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