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?
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* 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.
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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???
Yes, your calculations for the work done by the force seem to be correct.
To find the work done by a force, you need to take the dot product of the force vector and the displacement vector. In this case, the force vector is F = 3.1xi + 3.1j, and the displacement vector is the change in position, which is given as r2 - r1.
To calculate the work in the x-direction (W1), you need to integrate the x-component of the force with respect to x. As you correctly calculated, this gives you W1 = 30.38 J.
To calculate the work in the y-direction (W2), you need to integrate the y-component of the force with respect to y. As you correctly calculated, this gives you W2 = -19.84 J. The negative sign indicates that the force is doing work in the opposite direction of the displacement.
Finally, you can find the net work (Wnet) by adding W1 and W2. As you correctly calculated, this gives you Wnet = 11 J.
So, according to your calculations, the work done by the force is 11 J. Well done!