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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...