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A particle is displaced from (1, 2) m \; to \; (0, 0) m along the path y=2x^{3}. Work done by a force \vec{F}=\left ( x^{3}\hat{j}+y\hat{i} \right ) N acting on the particle during this displacement is

Option: 1

-1.5 \; J


Option: 2

-3.5 \; J


Option: 3

3 \; J

 


Option: 4

-2.5 \; J


Answers (1)

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Answer (1)

W=\int \vec{f}.\vec{dl}=\int x^{3}dy+\int ydx

       =\int_{2}^{0}\frac{y}{2}dy+\int_{1}^{0}2x^{3}dx

      =\frac{1}{2}\left [ \frac{y^{2}}{2} \right ]_{2}^{0}+2.\left [ \frac{x^{4}}{4} \right ]_{1}^{0}=-1.5\; J

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