Work Energy Power

JEE Main level, single correct, medium

25 of 183

Question 25 of 183

Consider a force F⃗=−xi^+yj^\vec{F}=-x\hat{i}+y\hat{j} . The work done by this force in moving a particle from point A(1,0)A\left(1,0\right) to B(0,1)B\left(0,1\right) along the line segment is : (all quantities are in SI units)

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Answer: (C) 11

It is given that a force F⃗=−xi^+yj^\vec{F}=-x\hat{i}+y\hat{j} acts on a particle. The particle is moved from point A(1,0)A\left(1,0\right) to B(0,1)B\left(0,1\right) along the line segment.
Work done by a variable force on the particle,
W=∫F⃗⋅dr⃗=∫F⃗⋅(dxi^+dyj^)W=\int \vec{F}\cdot \vec{dr}=\int \vec{F}\cdot \left(dx\hat{i}+dy\hat{j}\right)
(In two dimension, dr⃗=dxi^+dyj^\vec{dr}=dx\hat{i}+dy\hat{j}
and it is given F⃗=−xi^+yj^\vec{F}=-x\hat{i}+y\hat{j})

W=∫(−xi^+yj^)⋅(dxi^+dyj^)W=\int \left(-x\hat{i}+y\hat{j}\right)\cdot \left(dx\hat{i}+dy\hat{j}\right)
=∫−xdx+ydy=∫−xdx+∫ydy=\int -xdx+ydy=\int -xdx+\int ydy.
As the particle is displaced from A(1,0)A\left(1,0\right) to B(0,1)B\left(0,1\right), xx varies from 11 to 00 and yy varies from 00 to 11.
So, work will be,
W=−∫10xdx+∫01ydy=−[x22]10+[y22]01W=-{\int }_{1}^{0}xdx+{\int }_{0}^{1}ydy=-{\left[\dfrac{{x}^{2}}{2}\right]}_{1}^{0}+{\left[\dfrac{{y}^{2}}{2}\right]}_{0}^{1}
=12(0+(1)2+(1)2−0)=1J=\dfrac{1}{2}\left(0+{\left(1\right)}^{2}+{\left(1\right)}^{2}-0\right)=1J.

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