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Work Power and Energy question

2021 · 25 Jul · Shift 2 · Q69
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Work Power and Energy question

2021 · 25 Jul · Shift 2 · Q69

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A force of F = (5y + 20) j^\widehat jj​ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ‾\underline{\hspace{2cm}}​ J.
Numerical answer
View written solutionFree

Correct answer: 450

  1. Given force

The force is

F⃗=(5y+20) j^  N\vec F = (5y+20)\,\hat j \;\text{N}F=(5y+20)j^​N

Since the force is along the yyy-direction and the particle moves from y=0y=0y=0 to y=10y=10y=10, the displacement element is

dr⃗=dy j^d\vec r = dy\,\hat jdr=dyj^​
  1. Work done by a variable force

Work is given by

W=∫F⃗⋅dr⃗W = \int \vec F \cdot d\vec rW=∫F⋅dr

Substituting:

W=∫010(5y+20)j^⋅(dy j^)W = \int_0^{10} (5y+20)\hat j \cdot (dy\,\hat j)W=∫010​(5y+20)j^​⋅(dyj^​)

Since j^⋅j^=1\hat j \cdot \hat j = 1j^​⋅j^​=1,

W=∫010(5y+20) dyW = \int_0^{10} (5y+20)\,dyW=∫010​(5y+20)dy
  1. Evaluate the integral
W=[5y22+20y]010W = \left[\frac{5y^2}{2} + 20y\right]_0^{10}W=[25y2​+20y]010​

At y=10y=10y=10,

5(10)22+20(10)=5002+200=250+200=450\frac{5(10)^2}{2} + 20(10) = \frac{500}{2} + 200 = 250 + 200 = 45025(10)2​+20(10)=2500​+200=250+200=450

At y=0y=0y=0,

000

Therefore,

W=450−0=450 JW = 450 - 0 = 450\text{ J}W=450−0=450 J
  1. Final answer

The required integer is

450\boxed{450}450​
  1. Comparison with stored answer

Stored correct answer = 450450450

This matches the derived answer.

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