# Work and Force. Double Integration?

• Idoke
In summary, the problem at hand involves a particle in a path limited to the XY plane, acted upon by a force described by the function F(x,y)=9y \hat{\bf x}+7x \hat{\bf y} (*The numerical coefficients are in kg/s^2 ). The task is to find the work done by the force along different paths between points A = (0,0) [m] and B = (1,2) [m], including: 1) First along the X axis and then along the Y axis, 2) On the curve y=2x^2, and 3) On the curve y=\sqrt{2x}. The solution involves writing the path in
Idoke

## Homework Statement

A particle in a path limited to the XY plane is acted upon by a force described in the following function:
$${\bf F}(x,y)=9y \hat{\bf x}+7x \hat{\bf y}$$

(*The numerical coefficients are in $$kg/s^2$$)
What is the work that the force does along these paths, between the points A = (0,0) [m] and B = (1,2) [m]:

1. First along the X axis and then along the Y axis.
2. On the curve $$y=2x^2$$
3. On the curve $$y=\sqrt{2x}$$

## Homework Equations

$$W_C = \int_{C} \bold{F} \cdot \mathrm{d}\bold{s}$$

## The Attempt at a Solution

If this would be a one variable function of the force I would calculate the definite integral between the two points. But I am stumped by two things:
1. The function of the force is a two variable function, which I haven't learned to integrate, reading "Multiple Integers" on wikipedia got me more confused.
2. I know this is probably a non-conservative force and so the path does matter, but I don't know how to account for that in my calculations.

I feel like I should know this and it's getting kind of frustrating, but I'm sure someone could help because it all seems kind of basic.

Thanks.

Why don't you try to write a path in vector form, and then multiply it with force.

Hey,
What do you mean writing the path in vector form?
For the first question I get:
A) along the X axis:
Fd=9y (the force in the X direction) and y=0 so W=0.
B) along the Y axis:
Fd=7x. and of course the change in the x direction is again 0 (because i am moving on the y axis) so it's 0.
0+0=0 (or so I hear :) )

Is this right? and what do I do on the two other parts?

yes, that's right.

you can write ds = xdx + ydy and then plug it to work formula.

The problem asks the work when the point moves from (0,0) to (1,2). The first question refers to a rectangular path, with one piece coincident with an axis, the other parallel with the other one. The formulation is not clear, but it is certainly meant in this way. If the point moves along the axes only, it never reaches from (0.0) to (1,2).

The work along the first path is

$$W= \int _{(0,0)}^{(1,0)}{F_xdx}+\int_{(1,0)}^{(1,2)}{F_ydy}$$.

If you start the path along the y axis,

$$W= \int _{(0,0)}^{(0,2)}{F_ydy}+\int_{(0,2)}^{(1,2)}{F_xdx}$$

When you integrate along a line given with a Y(x) function, you have to use dy=dy/dx*dx in the integrand.

If y=x2, dy/dx = 2x,

$$W= \int _{0}^{1}{(18x^2 +7x*2x)dx}$$

ehild

Thank you!
Your method did work, although I still don't know exactly how, I understand that I will learn it soon so I guess I'll be OK.

Both of you have been immensely helpful,
Cheers.

## 1. What is work and how is it measured?

Work is a scientific concept that measures the amount of energy used to move an object. It is calculated by multiplying the force applied to an object by the distance it is moved in the direction of the force.

## 2. What is force and how is it related to work?

Force is a physical quantity that describes the push or pull on an object. It is represented by the symbol F and is measured in Newtons (N). Work is directly related to force, as it is the product of force and distance.

## 3. What is double integration and how is it used in relation to work and force?

Double integration is a mathematical technique used to find the area under a curve. In the context of work and force, it is used to calculate the work done by a varying force over a distance. It involves integrating the force function twice to find the work done.

## 4. What are the units of work and force?

The units of work are joules (J) and the units of force are Newtons (N).

## 5. Can work and force be negative?

Yes, work and force can be negative. This happens when the direction of the force and the displacement of the object are in opposite directions, resulting in negative work. Negative work indicates that energy is being taken away from the object rather than being added to it.

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