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Work Power and Energy question

2023 · 1 Feb · Shift 2 · Q71
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Work Power and Energy question

2023 · 1 Feb · Shift 2 · Q71

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A force F=(5+3y2)\mathrm{F}=\left(5+3 y^{2}\right)F=(5+3y2) acts on a particle in the yyy-direction, where F\mathrm{F}F is in newton and yyy is in meter. The work done by the force during a displacement from y=2 my=2 \mathrm{~m}y=2 m to y=5 my=5 \mathrm{~m}y=5 m is ‾\underline{\hspace{2cm}}​ J.
Numerical answer
View written solutionFree

Correct answer: 132

  1. Given force

    The force acts along the yyy-direction and depends on position: F(y)=5+3y2F(y)=5+3y^2F(y)=5+3y2

  2. Work done by a variable force

    Since force and displacement are along the same direction, W=∫y=25F(y) dyW=\int_{y=2}^{5} F(y)\,dyW=∫y=25​F(y)dy

    Substituting F(y)F(y)F(y): W=∫25(5+3y2) dyW=\int_{2}^{5} (5+3y^2)\,dyW=∫25​(5+3y2)dy

  3. Integrate

    ∫(5+3y2) dy=5y+y3\int (5+3y^2)\,dy = 5y + y^3∫(5+3y2)dy=5y+y3

    Therefore, W=[5y+y3]25W=\left[5y+y^3\right]_{2}^{5}W=[5y+y3]25​

  4. Apply limits

    At y=5y=5y=5: 5(5)+53=25+125=1505(5)+5^3=25+125=1505(5)+53=25+125=150

    At y=2y=2y=2: 5(2)+23=10+8=185(2)+2^3=10+8=185(2)+23=10+8=18

    Hence, W=150−18=132W=150-18=132W=150−18=132

  5. Final answer

    132 J\boxed{132\ \text{J}}132 J​

  6. Comparison with stored answer

    Stored correct answer = 132132132

    My derived answer matches the stored answer.

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