Physics/Math
posted by Technoboi11
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 = 30.3819.84
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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