Calculating work with force and position vectors

[Fizic]

Homework Statement

What work is done by a force [itex]\vec{F}[/itex]=((2 N/m)x)[itex]\hat{i}[/itex] + (3 N)[itex]\hat{j}[/itex], with x in meters, that moves a particle from a position [itex]\vec{r}[/itex]₁=(2 m)[itex]\hat{i}[/itex] + (3 m)[itex]\hat{j}[/itex] to a position [itex]\vec{r}[/itex]₂=-(4 m)[itex]\hat{i}[/itex] - (3 m)[itex]\hat{j}[/itex]

Homework Equations

Work=F[itex]\bullet[/itex]d

The Attempt at a Solution

[itex]\int_{2}^{-4}[/itex](2x)dx + [itex]\int_{3}^{-3}[/itex](3)dx

##\left.x^2\right|_2^{-4} + \left.3x\right|_3^{-3}##

##(16-4)+(-9-9)=-6 J##

Thanks.

 

[anathema]

Thank you for your post. It seems like you are on the right track with your solution. However, there are a few errors in your attempt.

Firstly, when calculating the work done by a force, we use the dot product between the force and displacement vectors. So the correct equation is W = F · d.

Secondly, the displacement vector is given by the difference between the final and initial positions, so it should be d = r2 - r1 = (-4 m - 2 m) i + (-3 m - 3 m) j = (-6 m) i + (-6 m) j.

Finally, when calculating the dot product, we use the magnitude of the force and the magnitude of the displacement vector, so the correct equation is W = |F| |d| cosθ, where θ is the angle between the two vectors. In this case, since the force and displacement vectors are in the same direction, the angle between them is 0 and cosθ = 1.

Putting all of this together, we get W = (|F|)(|d|)(cosθ) = (|F|)(|-6 m|)(1) = (5 N)(6 m) = 30 J.

I hope this helps clarify the solution for you. Let me know if you have any further questions or concerns.