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

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

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

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

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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.The force doesn't have to be the cause of the displacement. The cause is irrelevant for the definition of work.

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I'm sure you meant the component of forceWe're interested in the component of force that pointsin the same directionas the motion

What about the work involved in decelerating an object?

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$$m \dot{\vec{v}}=\vec{F}.$$

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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