A force 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) [2025]

Question

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) [2025]

Smart Achievers · Accepted Answer

(152)

y = 10 – x

$W = \int_{0}^{4} x^{2} (10 - x) d x + \int_{0}^{2} y^{2} d y$

$= {[\frac{10 x^{3}}{3} - \frac{x^{4}}{4}]}_{0}^{4} + {[\frac{y^{3}}{3}]}_{0}^{2} = \frac{640}{3} - 64 + \frac{8}{3} = 152$ .

Solution

Q.
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) [2025]

Ans.
(152)

y = 10 – x

$W = \int_{0}^{4} x^{2} (10 - x) d x + \int_{0}^{2} y^{2} d y$

$= {[\frac{10 x^{3}}{3} - \frac{x^{4}}{4}]}_{0}^{4} + {[\frac{y^{3}}{3}]}_{0}^{2} = \frac{640}{3} - 64 + \frac{8}{3} = 152$ .

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Solution

Q. A force f=x2yi^+y2j^ 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) [2025]

Ans. (152) y = 10 – x W=∫04x2(10–x)dx+∫02y2dy =[10x33–x44]04+[y33]02=6403–64+83=152.

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Q.
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) [2025]

Ans.
(152)

y = 10 – x

$W = \int_{0}^{4} x^{2} (10 - x) d x + \int_{0}^{2} y^{2} d y$

$= {[\frac{10 x^{3}}{3} - \frac{x^{4}}{4}]}_{0}^{4} + {[\frac{y^{3}}{3}]}_{0}^{2} = \frac{640}{3} - 64 + \frac{8}{3} = 152$ .