- #1

- 8

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Phenomniverse
- Start date

- #1

- 8

- 0

- #2

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

Work is the dot product of force and displacement vectors i.e. W=∫FWork = force x displacement

- #3

- 8

- 0

Again, I'm not sure why this means that displacement should be treated as a scalar when calculating work. Please clarify this for me if you can.Work is the dot product of force and displacement vectors i.e. W=∫F→.dx→. or |W|=F*x*cosθ. So, it is a scalar quantity.

- #4

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

It shouldn't. Force and displacement are both vectors. Their dot product i.e. 'work' is a scalar quantity. All the dot products give scalar quantities. Hence, dot product is also called 'scalar product'.displacement should be treated as a scalar

This means work is the product of|W|=F*x*cosθ

- #5

- 8

- 0

- #6

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

- #7

- 8

- 0

What is a conservative field?

- #8

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

Conservative field is a field in which work done is path independent. No matter how you go from A to B, work done is same along all the paths. Examples are gravitational field and electrostatic field. If you threw a ball up in the air with some veocity, it will go up and stop at some point. Then it will start falling and when it reaches the ground, it will have the same velocity as it had when it was thrown up. This means, work done on the ball is 0. Here, displacement of the ball changes its direction but force doesn't. Gravitational force always pulls the ball downward. Going up, the ball loses energy and falling back, it gains energy. Thus, net work done is 0.What is a conservative field?

- #9

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

In your 'walking across the room' example, "net displacement" is 0. But work is not Force*net displacement. It is actually ∫F⋅dx i.e. summation of force*instantaneous displacement between two points. Also, force and displacement both change their directions when you walk back to where you started.* Zero displacement does not necessarily mean zero work.*

Last edited:

- #10

- 8

- 0

- #11

- 18

- 1

U start pushing the box. U push the box a nanometer. stop. find the work by taking the (dot) product of force and displacement. U again push the box 1 nanometer .stop. Again find the work done .you need to do this an infinite number of times to find the total work done by adding the individual work for each moment.

Thus your work doesn't depend on the 'net displacement ' . Rather, it depends on the individual forces and displacements at every moment.

- #12

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,143

You are welcome!

When work is done, there is

Good luck with your studies!

- #13

- 183

- 15

for every step take take friction is always pushes you forward , so instantaneous displacement is in dir of instantaneous force.he net displacement is zero, however I am told that there is still work performed.

so instantaneous work is +ve , so net work by friction on body is non-zero

- #14

nasu

Gold Member

- 3,776

- 433

Who told you that the work is not zero and in what context?

You can calculate work as force time overall displacement only if the force is constant. And constant means both in magnitude and direction. Otherwise you nedd to use the integral form.

If you apply a constant force to an object moving from A to B, and then at B you switch the direction of the force at B and apply the force from B to A, the force is not the same. You need to integrate (or here, just to split into two). And the work done by this (or these two) force(s) is not zero.

If the direction of the force is not switched at B, then the work done by that force on the ABA path is indeed zero.

Context is important.

All this does not apply in a simple way to walking. Even if you stand still, your muscles do some work and you spend energy. It's better to start with simpler systems than a living being.

- #15

- 8

- 0

You can calculate work as force time overall displacement only if the force is constant.

Ok that's the critical information that I was missing. So if the force is changing you need to calculate the sum of all the forces and multiply that by the sum of all the displacements produced by those forces?

How does this relate to the explanation in terms of dot products and cross products?

The context was a biomechanics prac class. Hence the use of a living system as an example.

- #16

nasu

Gold Member

- 3,776

- 433

You add the dot products of force and displacements. Which for continuously varying force means do an integral.

- #17

- 8

- 0

I don't know what 'do an integral' means. Sounds like a skateboard trick of some sort.

- #18

nasu

Gold Member

- 3,776

- 433

But until you learn some more math you can divide the motion into portions so that the force is constant for each one and add the dot products (force times displacement) to get the total work.

An integral is something like this but where the portions are very small (infinitesimal).

Share:

- Replies
- 11

- Views
- 2K

- Replies
- 2

- Views
- 3K