# Work & Displacement: Scalar vs Vector

• Phenomniverse
In summary, work is the dot product of force and displacement vectors, making it a scalar quantity. In the example of walking across a room and back, the net displacement may be zero, but work is still being performed because force and displacement change direction at each moment. Work is not dependent on net displacement, but on the individual forces and displacements at each moment. When work is done, there is always a change in energy, and a body loses or gains energy depending on the direction of the force.

#### Phenomniverse

Work = force x displacement, so if there is zero displacement, the work is zero, right? But if I get up and walk across the room and walk back again to my starting place, the net displacement is zero, however I am told that there is still work performed. The explanation given for this is that work is a scalar quantity even though displacement is a vector. I'm not sure why this means that the net displacement needs to be broken into its component parts (effectively treating it as the scalar quantity of distance) in order to calculate the work performed...?

Phenomniverse said:
Work = force x displacement
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.

cnh1995 said:
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.
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.

Phenomniverse said:
displacement should be treated as a scalar
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'.
cnh1995 said:
|W|=F*x*cosθ
This means work is the product of displacement and component of force in the direction of displacement.

Can you explain this in the context of my example of walking across the room and back? Is the displacement zero, or isn't it? Is it realistic to talk about work in this example, or does the (apparent) zero quantity of displacement mean there is zero work performed?

Suppose you pushed a box from A to B with some force F. Work done by you in this process is F*d. If you pushed the block back to A, work done is again F*d. Note that while pushing the block back from B to A, both force and displacement changed their directions, so the work is not 0. If this were a conservative field, work done would be 0.

What is a conservative field?

Phenomniverse said:
What is a conservative field?
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.

Phenomniverse
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.

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Phenomniverse
Ok well I'm still far from being 100% clear on any of this but I appreciate your attempts to illuminate me. I'm new to studying physics and obviously struggling to grasp basic concepts. Perhaps I'll eventually gather enough knowledge to understand the rationale behind those concepts. Thanks for your help cnh1995.

So when you find the work done when u push the box from A to B and back, you really need to consider every moment.

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.

Phenomniverse
Phenomniverse said:
Ok well I'm still far from being 100% clear on any of this but I appreciate your attempts to illuminate me. I'm new to studying physics and obviously struggling to grasp basic concepts. Perhaps I'll eventually gather enough knowledge to understand the rationale behind those concepts. Thanks for your help cnh1995.
You are welcome!
When work is done, there is always change in energy of the body. When you move across the room and come back, your energy is lost. You do not gain any energy in this process. This means some non-zero work has been done. When work done is zero, initial and final energies of the body must be equal. If a body loses energy when going from A to B but gains the lost energy while coming back from B to A, work done is zero. Because here, initial energy and final energy are equal. That's what I was trying to say in #8 using the example of a ball thrown up against gravity. Going up, it loses energy and coming back, it regains energy. This happens because force doesn't change its direction, only displacement does. Hence, when the ball is going up, force of gravity is resisting its motion and the ball is losing energy. When the ball is falling down, the same force of gravity now aids its motion, hence, the ball gains energy.

Phenomniverse
Phenomniverse said:
he net displacement is zero, however I am told that there is still work performed.
for every step take take friction is always pushes you forward , so instantaneous displacement is in dir of instantaneous force.
so instantaneous work is +ve , so net work by friction on body is non-zero

Unfortunately you picked up a quite complex system, that of a man walking. This does not help your understanding.
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.

nasu said:
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.

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

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

No, it is not a skateboard trick but a calculus one. :)

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).

Phenomniverse

## What is work?

Work is defined as the amount of force applied over a distance. In scientific terms, work is equal to the product of the force and the displacement of an object in the direction of the force.

## What is displacement?

Displacement is the change in position of an object in a specific direction. It is measured in meters (m) in the SI system. Displacement can be either positive or negative, depending on the direction of movement.

## What is the difference between scalar and vector quantities?

Scalar quantities are those that have only magnitude, such as speed or temperature. On the other hand, vector quantities have both magnitude and direction, such as displacement or force. Work is a vector quantity as it has both magnitude (force) and direction (displacement).

## How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance it is displaced in the direction of the force. This can be represented mathematically as W = F*d, where W is work, F is force, and d is displacement.

## What are some real-life examples of work and displacement?

Some examples of work and displacement in everyday life include pushing a shopping cart, lifting weights at the gym, and pushing a lawn mower. In all of these examples, a force is applied to an object, causing it to move in a certain direction, resulting in displacement and work being done.