First slide

Work done by Diff type of forces

Question

A force $\bar{F} = (y^{2} - x^{2} + z^{2}) \hat{i} + (3 xy - 5 z) \hat{j} + 4 z \hat{k}$ is applied on a particle from the point (0,0,0) to the point (2,4,0) in the path shown . If $W_{A}$ is the work done by the force in the paths A, then $W_{A}$ =?

Moderate

A

40 J
B

$\frac{136}{3} J$
C

$\frac{632}{35} J$
D

Zero

Solution

The work done by the external force is $\int \vec{F} . \vec{dS}$ . Now here in this case the z coordinate of the motion of the particle is zero. $\vec{F} = (y^{2} - x^{2}) \hat{i} + 3 xy \hat{j}$
Hence $W_{A} = \int_{0}^{2} - x^{2} dx + \int_{0}^{4} 3xy dy = {[\frac{- x^{3}}{3}]}_{0}^{2} + {[3 y^{2}]}_{0}^{4} = \frac{- 8}{3} + 48 = \frac{136}{3}$

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