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

  • question_answer
    Calculate the work done on the tool by \[\vec{F}\left( 11.25\hat{i}+11.25\hat{j} \right)N\] if the tool is first moved out along the x-axis to the point \[x=3.00m,\] \[y=0\] and then moved parallel to the y-axis to\[x=3.00m,\]\[y=3.00\text{ }m\].

    A) 67.5 J

    B) 85 J  

    C) 102 J

    D) 7.5 J

    Correct Answer: A

    Solution :

    [a] Net displacement is \[\left( 3 \hat{i}+3\hat{j} \right)\] so work done \[= \vec{F}.\Delta \vec{s} = 11.25\left( \hat{i}+\hat{j} \right).\left( 3 \hat{i}+ 3\hat{j} \right)\] \[= 33.75 \times  2=67.50 J\]


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