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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Calculation of workEasy2 minQB

PhysicsEasynumerical

A force $f = x^{2} y \hat{i} + y^{2} \hat{j}$ acts on a particle in a plane $x + y = 10$ . The work done by this force during a displacement from $(0, 0)$ to $(4 m, 2 m)$ is Joule (round off to the nearest integer)

Answer:

152.00

Solution:

$y = 10 - x w = \int_{0}^{4} x^{2} (10 - x) d x + \int_{0}^{2} y^{2} d y = \frac{10 x ^{3}}{3} - \frac{x ^{4}}{4}_{0}^{4} + \frac{y ^{3}}{3}_{0}^{2} = \frac{640}{3} - \frac{256}{4} + \frac{8}{3} = 216 \times 64 = 152 J$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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