Calculating work with force and position vectors

In summary: Your Name]In summary, the work done by the force F=((2 N/m)x)i + (3 N)j, with x in meters, on a particle moving from position r1=(2 m)i + (3 m)j to position r2=-(4 m)i - (3 m)j is 30 J. This is calculated using the dot product between the force and displacement vectors, which are (|-6 m|) and 1, respectively. The angle between these vectors is 0, resulting in a cosθ value of 1.
  • #1
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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.
 
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  • #2


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.


 

FAQ: Calculating work with force and position vectors

What is work in physics?

Work in physics is defined as the product of force and displacement. It is calculated by multiplying the magnitude of the force applied in the direction of the displacement by the distance over which the force is applied.

How do you calculate work with force and position vectors?

To calculate work with force and position vectors, you first need to determine the magnitude of the force and the displacement. Then, you can use the formula W = F * d * cos(theta), where F is the force vector, d is the displacement vector, and theta is the angle between the two vectors.

What is the unit of measurement for work?

The unit of measurement for work is joules (J). This is equivalent to a newton-meter (N*m) in the SI system.

Can work be negative?

Yes, work can be negative. This occurs when the force and displacement vectors are in opposite directions, resulting in a negative value for work. This indicates that the force is acting in the opposite direction of the displacement.

How is work related to energy?

Work and energy are closely related concepts in physics. The work done on an object is equal to the change in its energy. This is known as the work-energy theorem. In other words, when work is done on an object, it gains or loses energy depending on the direction of the force and displacement vectors.

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