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

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Following the official treatment: $W=\int_0^4x^2y\,dx+\int_0^2y^2\,dy$ with $y=10-x$ in the first term. $\int_0^4x^2(10-x)\,dx=\frac{640}3-64=\frac{448}3$ and $\int_0^2y^2\,dy=\frac83$. $W=\frac{448+8}3=152$ J. (Medium confidence: the end point $(4,2)$ does not lie on $x+y=10$, so the problem is loosely posed; this follows the key.) Answer: 152.

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

Explanation

Reviewed by Dr. Abubacker

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