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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Work Done by Constant Force

  • question_answer
    A force of 5 N, making an angle \[\theta \] with the horizontal, acting on an object displaces it by \[0.4m\] along the horizontal direction. If the object gains kinetic energy of 1J, the horizontal component of the force  is [EAMCET (Engg.) 2000]

    A)             1.5 N  

    B)             2.5 N

    C)             3.5 N  

    D)             4.5 N

    Correct Answer: B

    Solution :

                    Work done on the body = K.E. gained by the body             \[FS\cos \theta =1\ Joule\]Þ\[F\cos \theta \]\[=\frac{1}{s}=\frac{1}{0.4}=2.5N\]


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