WOrk done by a variable force (in two dimension)

In summary: I stated before, the particle goes from (x1,y1) to (x2, y2)so the end point is (x2, y2)So, am I doing this right?P/s: however, the path between the two points is straight line
  • #1
daivinhtran
68
0

Homework Statement



How to find the work done by a variable force in (two dimension)

When
F = ax^2 i + b y^3 j

If a subject move from (x1,y1) to (x2, y2)

Homework Equations



F = dW/dr

The Attempt at a Solution



I tried to solve them separately by x-direction and y-direction, and then I added them up
like Wx + Wy = W
Am I right?
 
Last edited:
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  • #2
In this problem the path taken from (x1,y1) to (x2, y2) is important, different paths taken give different results. In this problem I assume the path is a straight line. If you solve the problem as you suggested I'm not sure you took into account the path taken? Can you scan your work and post it?

Hopefully others will help.
 
  • #3
Spinnor said:
In this problem the path taken from (x1,y1) to (x2, y2) is important, different paths taken give different results. In this problem I assume the path is a straight line. If you solve the problem as you suggested I'm not sure you took into account the path taken? Can you scan your work and post it?

Hopefully others will help.


https://www.physicsforums.com/attachment.php?attachmentid=51764&stc=1&d=1349922834

I hope I solve it right
 

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  • #4
I don't think what you did is right. You did not take the path into account only the end points. Start from the diffinition of infitesimal work, dW = vector F dot vector dr and plug and chug. You must come up with a vector expression for dr for the straight line path. You can do this as you know the endpoints.
 
  • #5
Spinnor said:
I don't think what you did is right. You did not take the path into account only the end points. Start from the diffinition of infitesimal work, dW = vector F dot vector dr and plug and chug. You must come up with a vector expression for dr for the straight line path. You can do this as you know the endpoints.

as I stated before, the particle goes from (x1,y1) to (x2, y2)
so the end point is (x2, y2)

So, am I doing this right?

P/s: however, the path between the two points is straight line
 
  • #6
I don't think so. Hopefully others will confirm.
 

FAQ: WOrk done by a variable force (in two dimension)

What is work done by a variable force?

Work done by a variable force is the measure of energy transfer that occurs when a force is applied to an object and causes it to move in a two-dimensional plane.

How is work calculated for a variable force in two dimensions?

Work is calculated by multiplying the magnitude of the force by the displacement of the object in the direction of the force. This can be represented by the equation W = F * d * cos(theta), where theta is the angle between the force and displacement vectors.

What is the difference between work done by a constant force and a variable force in two dimensions?

Work done by a constant force only considers the magnitude of the force, while work done by a variable force takes into account the direction of the force and the displacement. This means that for a variable force, the work done may be positive, negative, or zero depending on the angle between the force and displacement vectors.

Can work done by a variable force be negative?

Yes, work done by a variable force can be negative. This occurs when the angle between the force and displacement vectors is greater than 90 degrees, resulting in a negative value for the cosine term in the work equation.

How does the graph of work done by a variable force in two dimensions look like?

The graph of work done by a variable force in two dimensions is typically a curved line, as the work done changes with the angle between the force and displacement vectors. The maximum work done occurs when the force and displacement vectors are parallel, and the minimum (or negative) work is done when they are perpendicular.

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