## Monday, January 14, 2019

### Unizor - Physics4Teens - Energy - Kinetic Energy - Introduction

Notes to a video lecture on http://www.unizor.com

Kinetic Energy - Introduction

Kinetic energy is a quantitative characteristic of an object's motion,
that signifies that this motion can result in some work, when our object
interacts with surrounding environment.

As an example, consider an object of mass m uniformly moving within some inertial frame along a straight line trajectory with speed v.

Assume that at some moment of time there appears a constant force F that acts against its motion. For example, a constant air resistance.

As a result of this interaction with air molecules, our object will slow
down because of air resistance acting against its motion until it
stops. So, the motion of our object caused some work - moving molecules
of air away from the trajectory, after which the motion of our object no
longer exists, it stops completely.

Obviously, instead of air resistance, we can consider friction or
gravity, or any other force, assumed constant for this experiment.

Let's calculate the work done by this force, as it acts against the motion of our object and causes its deceleration.

First of all, we calculate the deceleration a:

a = F/m

Then, knowing deceleration, we get the time t our force acts until an object stops:

t = v/a = m·v/F

Now the distance S our object travels until a full stop:

S = a·t²/2 = F·t²/(2m) =

= F·m²·v²/(2m·F²) =

= m·v²/(2F)

Finally, the work W performed by our force F during the distance S:

W = F·S = m·v²/2

What's most remarkable about this formula is that the work performed by force F does not depend on this force, but only on the characteristic of an object (mass m) and its motion (speed v).

So, it appears that there is something specific for an object (its mass) and its motion (speed) - the amount of work to bring this object to a state of rest, and this quantity of work equals to m·v²/2.

This quantitative characteristic of an object in motion is called its kinetic energy.

Let's slightly complicate the problem. The constant force F slows down our object not to a full stop, but to final speed vend. What will be the work force F would perform?

As before,

a = F/m

t = (v−vend)/a = m·(v−vend)/F

S = v·t − a·t²/2 =

= m·v·(v−vend)/F −

− F·m²·(v−vend)²/(2m·F²) =

= m·v·(v−vend)/F −

− m·(v−vend)²/(2F) =

= (m/2F)·(v²−v²end)

Finally, the work W performed by our force F during the distance S:

W = F·S = m·(v²−v²end)/2 =

= m·v²/2 − m·v²end)/2

This formula indicates that the work needed to change the state of a moving object from a state with one value of its kinetic energy to another equals to a difference between the values of its kinetic energy at these two states.

Absolutely analogous calculations can prove that the work of a constant force that accelerates an object of mass m from a state of rest to speed v equals to

W = m·v²/2 for any force,

and the work needed to accelerated an object from speed vbeg to v equals to

W = m·v²/2 − m·v²beg/2

In other words, work, that results from the interaction of moving object with surrounding environment, and its kinetic energy are intimately related. One can be converted into another and vice versa.

Work performed by different forces is, by definition, additive.

It means that, if force F1 acted on a distance S1 performing work W1=F1·S1 and force F2 acted on a distance S2 performing work W2=F2·S2, then the total amount of work performed by both forces is a sum of their individual work.

Immediate consequence of this is that kinetic energy of a system of objects, each having its own mass and moving with its own speed, equals to a sum of their individual kinetic energies.

That is, kinetic energy is additive.

If N objects of masses m1, m2, ... mN are moving with speeds v1, v2, ... vN then their total kinetic energy equals to

W = Σi∈[1,N](mi·v²i)/2

As a final example, consider a case when an object of mass m initially moves along the X-axis. The X-component of its velocity vector is Vx and its Y-component of a velocity vector Vy is zero. Then its speed V (a scalar) equals to its X-component Vx (a scalar) and its kinetic energy is

Ek = m·V²/2

where V = Vx

Assume that the force F acts at an angle φ to the X-axis.

In this case we can represent the vector of force F as a sum of two perpendicular vectors of force: Fx acting along the X-axis and Fy acting along the Y-axis.

Obviously,

F = Fx + Fy

These two perpendicular to each other forces, applied to our object,
give it a vector of acceleration that also can be represented as a sum
of two perpendicular vectors

ax = Fx /m

ay = Fy /m

Let's assume that the force F acts for a duration of time t.

During this time the force F
performs certain work and the velocity of an object will change. We
will compare the amount of work performed by the force with a change in
the kinetic energy of an object.

Considering the initial speed of an object along the X-axis was Vx and acceleration ax, the distance covered by our object along the X-axis equals to

Sx = Vx·t + ax·t² /2

At the same time our object moved along the Y-axis with initial speed 0 and acceleration ay, covering the distance

Sy = ay·t² /2

In terms of components of the force F the distances along the coordinates are

Sx = Vx·t + Fx·t² /(2m)

Sy = Fy·t² /(2m)

The total work performed by force F , considering work is additive and can be summarized in each direction, is

W = Fx·Sx + Fy·Sy =

= Fx·Vx·t + Fx²·t²/(2m) +

+ Fy²·t²/(2m)

Now let's calculate the kinetic energy of an object at the end of time period t.

The X-component of the velocity will be equal to

Vxt = Vx + ax·t = Vx + Fx·t /m

The Y-component of the velocity will be equal to

Vyt = ay·t = Fy·t /m

The object's kinetic energy at the end of the time period t, equals to

Ekt = m·(Vt/2

where Vt is the speed at time t.
Using the Pythagorean Theorem, we can represent (Vt as

(Vt)² = (Vxt)² + (Vyt.
Therefore, the kinetic energy can be summarized from X- and Y-components and can be calculated as a sum of:

Ext = m·(Vxt/2

Eyt = m·(Vyt/2

The increment of the kinetic energy is

ΔE = Ext + Eyt − Ek

All we have to do is to compare amount of work W performed by force F and increment of kinetic energy ΔE and make sure that they are equal.

Indeed,

ΔE = m·(Vx+Fx·t /m)²/2 +

+ m·(Fy·t /m)²/2 − m·Vx²/2 =

= Fx·Vx·t + Fx²·t²/(2m) +

+ Fy²·t²/(2m) = W

Therefore, the work performed by a force acting on an object during
certain period of time equals to an increment of a kinetic energy of
this object
.