JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    A force F acts on a particle such that its position x changes as shown in the figure. The work done by the particle as it moves from \[x=0\] to 20 m is  

    A) 37.5 J

    B) 10 J

    C) 45 J

    D) 22.5 J  

    Correct Answer: C

    Solution :

    [c] \[\operatorname{W} = area of F - x graph\] \[=area\text{ }of\text{ }\Delta +area\text{ }of\text{ }rectangle+area\text{ }of\text{ }\Delta \] \[=\frac{5\times 3}{2}+10\times 3+\frac{5\times 3}{2}=45J\]


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