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Work done by a vector field

  • Thread starter Duncan1382
  • Start date
  • #1

Homework Statement


Find the work done by the force field F in moving an object from P to Q.

[tex]F(x,y,z)=10y^(3/2)i+15x\sqrt{y}j[/tex]

P(1,1), Q(2,9)


Homework Equations


W = [tex]\int[/tex]F dot dr


The Attempt at a Solution


I have no clue how to do it
 

Answers and Replies

  • #2
tiny-tim
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Hi Duncan1382! :smile:

(have an integral: ∫ and a square-root: √ :wink:)

Draw the line from P to Q. How long is it?

Define a parameter along it, and integrate force dot distance wrt that parameter. :wink:

(time-saving tip: if part of the force is conservative, you can ignore that part)
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement


Find the work done by the force field F in moving an object from P to Q.

[tex]F(x,y,z)=10y^(3/2)i+15x\sqrt{y}j[/tex]
I assume you mean
[tex]F(x,y,z)= 10y^{3/2}i+ 15x\sqrt{y}j[/tex]
Use { } to surround things you want in an exponent,

P(1,1), Q(2,9)


Homework Equations


W = [tex]\int[/tex]F dot dr


The Attempt at a Solution


I have no clue how to do it
You give [tex]W= \int F\cdot dr[/tex]
Do you not know what that means? [itex]dr= idx+ jdy+ kdz[/tex] take the dot product of that and [itex]10y^{3/2}i+ 15x\sqrt{y}j[/itex] and integrate. Of course, you should state what path is to be used. In general the integral of a function of several variables from one point to another depends upon the path between the two points as well as the points themselves, but, fortunately the given force vector function is "conservative" so it does not matter which path you choose. Tiny-tim suggests the simplest- the straight line between the two points.

But because the force function is conservative, it has a "potential" function. That is, there is a scalar function, f(x,y,z) such that
[tex]\nabla f= \frac{\partial f}{\partial x}i+ \frac{\partial f}{\partial y}j+ \frac{\partial f}{\partial z}= 10y^{3/2}i+ 15x\sqrt{y}j[/tex]

That is, you have
[tex]\frac{\partial f}{\partial x}= 10y^{3/2}[/tex]
[tex]\frac{\partial f}{\partial y}= 16xy^{1/2}[/tex]
[tex]\frac{\partial f}{\partial z}= 0[/tex]

You should be able to solve those for f (up to a constant), then evaluate between the two points to see how much the potential energy has change- which, of course, is the work done on the particle.
 

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