# Work, Energy Power

## Scalar Product

There are two ways of multiplying vectors. One way is called the scalar product, while another is called the vector product.

Let us take any two vectors A and B. Their scalar product (or dot product) is denoted as A.B (A dot B).

A.B = A B cos θ ………….(1)

In this case, since A, B and cos θ are scalars so the dot product is a scalar quantity.

Above equation can also be written as follows:

A.B = A(B cos θ)

= B(A cos θ)

This figure shows two vectors A and B at angle θ to each other. In the second figure, the blue line shows the projection of B onto A. The blue line (in the third figure) shows the projection of A onto B. So, A.B is the product of magnitude of A and the component of B along A. Similarly, it is also the product of the magnitude of B and the component of A along B.

The scalar product follows the commutative law

A.B = B.A

The scalar product obeys the distributive law.

A.(B + C) = A.B + A.C

Moreover, A.(λB)= λ(A.B)

Where, λ is a real number.

For unit vectors, **î, ĵ, k̂** we have

**î.ĵ = ĵ.ĵ = k̂.k̂** = 1

**î.ĵ = ĵ.k̂ = k̂.î** = 0

If there are two vectors as follows:

`A=A_x\î+A_y\ĵ+A_z\k̂`

`B=B_x\î+B_y\ĵ+B_z\k̂`

Then the scalar product is as follows:

`A.B=(A=A_x\î+A_y\ĵ+A_z\k̂).(B=B_x\î+B_y\ĵ+B_z\k̂)`

`=A_x\B_x+A_y\B_y+A_z\B_z`

So, we can also write

`A.A=A_x\A_x+A_y\A_y+A_z\A_z`

Or, `A^2=A_x^2+A_y^2+A_z^2`

If A and B are perpendicular then θ = 0°. As cos 90° = 0

So, in this case: A.B = 0

## Work-Energy Theorem

Let us recall the following equation of motion:

`v^2-u^2=2as` ………………(2)

Here, u is initial speed, v is final speed, a is acceleration and s is distance.

Multiplying both sides by `m/2`, we get

`1/2mv^2-1/2mu^2=mas`

Since `m×=F` so above equation can be written as follows:

`1/2mv^2-1/2mu^2=mas=Fs`

By employing vectors, equation (2) can be generalized as follows

`v^2-u^2=2a.d`

Here, a and d are acceleration and displacement vectors

Multiplying both sides by `m/2`we get

`1/2mv^2-1/mu^2=m\a.d=F.d` ……………(3)

We know

`1/2mv^2=K` (Kinetic energy)

`F.d=W` (Work)

So, equation (3) can be written as follows:

`K_f-K_i=W` ……………(4)

Here, K_{i} is initial kinetic energy, and K_{f} is final kinetic energy. In the context of this equation, work can be defined as follows:

*“The change in kinetic energy of a particle is equal to the work done on it by the net force.”*

## Work

Let us take an object which undergoes displacement d in the positive x-direction when a constant force F is acting on it. In this case, the product of component of force (in the direction of displacement) and the magnitude of displacement is called work done.

W = (F cos θ)d = F.d

No work is done if

- Displacement = 0
- Force = 0
- Force and displacement are in mutually perpendicular direction. (Because cos 90° = 0)
- If θ between 0° and 90° -- Work done is positive
- If θ between 90° and 180° -- Work done is negative

Work and energy have the same dimensions [ML^{2}T^{-2}]. SI units of work and energy are joule (J).

### Work Done by Variable Force

In real life, we rarely encounter constant force. Variable force is the norm in common situations.

If displacement Δx is small, the force F(x) can be taken as constant. In this case, work done is as follows:

ΔW = F(x) Δx

This graph shows variable force Vs displacement. Successive rectangles (in graph) give work done at particular instant of displacement. Total work is equal to the sum of areas of successive rectangles. This can be given by following equation.

W ∼ Σ F(x) Δx

#### Work-Energy Theorem for Variable Force

The time rate change of kinetic energy is

`(dK)/(dt)=d/(dt)(1/2mv^2)=m(dv)/(dt)v`

Or, `(dK)/(dt)=Fv`

Or, `(dK)/(dt)=F(dx)/(dt)`

So, `dK=Fd\x`

On integration,

Or,

We have earlier seen that K_{f} - K_{i} = W

So, WE (Work-Energy) theorem is proved for variable force.

## Potential Energy

Let us assume a ball with mass m is raised to a height h. Work done by the external agency against gravitational force is

`W = mgh`

The work done gets stored as potential energy. Gravitational potential energy is denoted by V(h); as a function of height h. It is the negative work done by gravitational force in raising the object to that height.

`V(h)=mgh`

The gravitational force F is equal to the negative of the derivative of V(h) with respect to h. Thus,

`F=-d/(dh)V(h)=-mg`

The negative sign shows the downward direction of gravitational force. When the ball is released, it come s down with increasing speed. The speed of ball (just before hitting the ground) is given by following equation.

`v^2=2gh`

This equation can be written as

`1/2mv^2=mgh`

This means that the gravitational potential energy of the object at height h, manifests as kinetic energy on reaching the ground.

The work done by a conservative force (such as gravity) depends on initial and final positions only. If the work done or kinetic energy did depend on other factors (such as velocity or a particular path taken by the object) the force would be called non-conservative.