A force ⃗F=x²y ̂i+y² ̂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).

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Accepted Answer

152. W=∫ x²y dx+∫ y² dy. Use the given relation y=10-x in the first integral, taken from x=0 to x=4: ∫₀⁴x²(10-x) dx=[(10x³)/3-(x⁴)/4]₀⁴=(640)/3-64=(448)/3. Second integral from y=0 to y=2: ∫₀²y² dy=8/3. W=(448)/3+8/3=(456)/3=152 J. Note: neither end point lies on x+y=10 and the force is not conservative, so the result depends on this substitution; it is the route the official key follows.

A force ⃗F=x²y ̂i+y² ̂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).

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