Posted by **Technoboi11** on Saturday, March 3, 2007 at 2:10pm.

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?

For Further Reading

* Physics/Math - bobpursley, Thursday, March 1, 2007 at 6:27am

Work is the dot product of Force and displacement.

Change in displacement is r2 minus r1. I will be happy to critique your work on this.

Rememeber dot product is a scalar, the sum of i component times i component plus j component times j component etc.

>>>>>>>>>>>>>>>>>>>>>>>>>>>

Here is what I did:

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

Is this correct???

The procedure you set up is correct. I don't have a calculator here, so I didn't check math.

What is wrong with my thinking here???

I got the right answer...stupid careless error on my behalf.

The only thing I see "wrong" is the 11 J is a rounded number.

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