Why work done by a force is a scalar product

saurabhjain

Why work done by a force was taken as dot product between force applied and displacement caused?

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axmls

Because 1. We're interested in the component of force that points in the same direction as the motion, and 2. It's scalar because energy is a scalar quantity (and work is a measure of energy).

Drakkith

Staff Emeritus
2018 Award
Because 1. We're interested in the component of force that points in the same direction as the motion
What about the work involved in decelerating an object?

A.T.

Why work done by a force was taken as dot product between force applied and displacement caused?
The force doesn't have to be the cause of the displacement. The cause is irrelevant for the definition of work.

• kbr2020

Chandra Prayaga

The force doesn't have to be the cause of the displacement. The cause is irrelevant for the definition of work.
Yes. In fact, there could be many forces acting on the body, and the body could be moving along a very complicated path because of all the forces. Each force does work, which is calculated using the standard dot product definition of work. In this process, each force could be doing positive, zero or negative work. The algebraic sum of the works done by all the forces, which we can call the net work, can also be calculated by first finding the net force, and then calculating the work by that net force.

Doc Al

Mentor
We're interested in the component of force that points in the same direction as the motion
I'm sure you meant the component of force parallel to the displacement. The force could very well be opposite to the displacement, which was Drakkith's point here:
What about the work involved in decelerating an object?

Chandra Prayaga

Again, if there is one force that is reducing the speed of an object (I don't use the word deceleration), then the direction of the force is opposite to the velocity, and the work done turns out to be negative. The definition of work is still the same, it is the dot product of force and displacement.

vanhees71

Gold Member
The reason for taking this specific functional is that it is useful. As the most simple example take a single point particle moving in some external force field. Then Newton's equation of motion reads
$$m \dot{\vec{v}}=\vec{F}.$$
Then you can multiply this with $\vec{v}$ (scalar product), and you immediately see that the left-hand side is a total time derivative:
$$m \vec{v} \cdot \dot{\vec{v}}=\frac{m}{2} \mathrm{d}_t (\vec{v}^2)=\mathrm{d}_t T=\vec{v} \cdot \vec{F}$$
with the kinetic energy
$$T=\frac{m}{2} \vec{v}^2.$$
Integrating this over a time interval you get
$$T_2-T_1=\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}.$$
This is known as the work-action theorem.

As it stands, it's not so useful, because you need to know the solution of the equation of motion to evaluate the integral. But very often forces in Newtonian mechanics are derivable as the gradient of a scalar field, i.e., a potential:
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Then the integrand becomes also a total time derivative:
$$\vec{v} \cdot \vec{F}=\dot{\vec{x}} \cdot \vec{F}=-\dot{\vec{x}} \cdot \vec{\nabla} V=-\mathrm{d}_t V.$$
Then the right-hand side of the work-action theorem becomes
$$T_2-T_1=-(V_2-V_1).$$
You don't need to know the trajectory of the particle anymore but only the positions and velocities of the particle at the initial and final time $t_1$ and $t_2$ of the motion. Writing the total energy as
$$E=T+V,$$
the above equation is nothing else than the energy-conservation law
$$E_2=E_1.$$
That's why forces which are the gradient of a time-independent scalar field are called "conservative", because they lead to energy conservation.

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