Work done by force moving a particle

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SUMMARY

The work done by the force F = 3.1xi + 3.1j while moving a particle from position r1 = 2.1i + 2.5j to r2 = -4.9i -3.9j is calculated using the integral of the force components. The calculations yield W1 = 30.38 J for the x-component and W2 = -19.84 J for the y-component, resulting in a net work Wnet = 11 J. However, the solution is deemed incorrect, prompting a reevaluation of whether the force F is conservative.

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  • Understanding of vector calculus
  • Knowledge of work-energy principles
  • Familiarity with conservative forces
  • Ability to perform definite integrals
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Homework Statement



What work is done by a force (in Newtons) F = 3.1xi + 3.1j, with x in meters, that moves a particle from a position r1 = 2.1i + 2.5j to r2 = - -4.9i -3.9j?

Homework Equations



Work=integral(Fdx)

The Attempt at a Solution



W1=integral of Fxdx = integral (3.1x)dx
= 1.55x^2 evaluated at x2=-4.9 and x1=2.1.
= [(1.55(-4.9)^2)-(1.55(2.1)^2)]
= 30.38 J
W2=integral of Fydy = integral (3.1)dy
= 3.1[(-3.9)-(2.5)]
= -19.84 J
Wnet = W1+W2
Wnet = 30.38-19.84
Wnet = 11 J

But my answer, 11 J, is incorrect...
 
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This is totally incorrect, on several levels.

To give you a hint of how to solve this properly, answer the following question:
Is F a conservative force?
 

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