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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...