*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

**uniformly moving within some inertial frame along a straight line trajectory with speed**

*m***.**

*v*Assume that at some moment of time there appears a constant force

**that acts against its motion. For example, a constant air resistance.**

*F*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

**our force acts until an object stops:**

*t*

*t = v/a = m·v/F*Now the distance

**our object travels until a full stop:**

*S*

*S = a·t²/2 = F·t²/(2m) =*

= F·m²·v²/(2m·F²) =

= m·v²/(2F)= F·m²·v²/(2m·F²) =

= m·v²/(2F)

Finally, the work

**performed by our force**

*W***during the distance**

*F***:**

*S*

*W = F·S = m·v²/2*What's most remarkable about this formula is that the work performed by force

**does not depend on this force, but only on the characteristic of an object (mass**

*F***) and its motion (speed**

*m***).**

*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

**slows down our object not to a full stop, but to final speed**

*F***. What will be the work force**

*v*_{end}**would perform?**

*F*As before,

*a = F/m*

*t = (v−v*_{end})/a = m·(v−v_{end})/F

*S = v·t − a·t²/2 =*

= m·v·(v−v

− F·m²·(v−v

= m·v·(v−v

− m·(v−v

= (m/2F)·(v²−v²= m·v·(v−v

_{end})/F −− F·m²·(v−v

_{end})²/(2m·F²) == m·v·(v−v

_{end})/F −− m·(v−v

_{end})²/(2F) == (m/2F)·(v²−v²

_{end})Finally, the work

**performed by our force**

*W***during the distance**

*F***:**

*S*

*W = F·S = m·(v²−v²*

= m·v²/2 − m·v²_{end})/2 == m·v²/2 − m·v²

_{end})/2This 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

**from a state of rest to speed**

*m***equals to**

*v***for any force,**

*W = m·v²/2*and the work needed to accelerated an object from speed

**to**

*v*_{beg}**equals to**

*v*

*W = m·v²/2 − m·v²*_{beg}/2In 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

**acted on a distance**

*F*_{1}**performing work**

*S*_{1}**and force**

*W*_{1}=F_{1}·S_{1}**acted on a distance**

*F*_{2}**performing work**

*S*_{2}**, then the total amount of work performed by both forces is a sum of their individual work.**

*W*_{2}=F_{2}·S_{2}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

**objects of masses**

*N***,**

*m*_{1}**, ...**

*m*_{2}**are moving with speeds**

*m*_{N}**,**

*v*_{1}**, ...**

*v*_{2}**then their**

*v*_{N}**total kinetic energy**equals to

*W =*Σ_{i∈[1,N]}

**(**

*m*

_{i}**·v²**_{i}**)/**

*2*As a final example, consider a case when an object of mass

**initially moves along the X-axis. The X-component of its velocity vector is**

*m***and its Y-component of a velocity vector**

*V*_{x}**is zero. Then its speed**

*V*_{y}**(a scalar) equals to its X-component**

*V***(a scalar) and its**

*V*_{x}**kinetic energy**is

*E*_{k}= m·V²/2where

*V = V*_{x}Assume that the force

**acts at an angle**

*F***to the X-axis.**

*φ*In this case we can represent the vector of force

**as a sum of two perpendicular vectors of force:**

*F***acting along the X-axis and**

*F*_{x}**acting along the Y-axis.**

*F*_{y}Obviously,

**=**

*F***+**

*F*_{x}

*F*_{y}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

*a*_{x}= F_{x}/m

*a*_{y}= F_{y}/mLet's assume that the force

**acts for a duration of time**

*F***.**

*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

**and acceleration**

*V*_{x}**, the distance covered by our object along the X-axis equals to**

*a*_{x}

*S*_{x}= V_{x}·t + a_{x}·t² /2At the same time our object moved along the Y-axis with initial speed

**and acceleration**

*0***, covering the distance**

*a*_{y}

*S*_{y}= a_{y}·t² /2In terms of components of the force

**the distances along the coordinates are**

*F*

*S*_{x}= V_{x}·t + F_{x}·t² /(2m)

*S*_{y}= F_{y}·t² /(2m)The total work performed by force

**, considering work is additive and can be summarized in each direction, is**

*F*

*W = F*

= F

+ F_{x}·S_{x}+ F_{y}·S_{y}== F

_{x}·V_{x}·t + F_{x}²·t²/(2m) ++ F

_{y}²·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

*V*_{x}^{t}= V_{x}+ a_{x}·t = V_{x}+ F_{x}·t /mThe Y-component of the velocity will be equal to

*V*_{y}^{t}= a_{y}·t = F_{y}·t /mThe object's kinetic energy at the end of the time period

**, equals to**

*t*

*E*_{k}^{t}= m·(V^{t})² /2where

**is the speed at time**

*V*^{t}**.**

*t*Using the Pythagorean Theorem, we can represent

**as**

*(V*^{t})²**.**

*(V*^{t})² = (V_{x}^{t})² + (V_{y}^{t})²Therefore, the kinetic energy can be summarized from X- and Y-components and can be calculated as a sum of:

*E*_{x}^{t}= m·(V_{x}^{t})² /2

*E*_{y}^{t}= m·(V_{y}^{t})² /2The increment of the kinetic energy is

Δ

*E = E*_{x}^{t}+ E_{y}^{t}− E_{k}All we have to do is to compare amount of work

**performed by force**

*W***and increment of kinetic energy Δ**

*F***and make sure that they are equal.**

*E*Indeed,

Δ

*E = m·(V*

+ m·(F

= F

+ F_{x}+F_{x}·t /m)²/2 ++ m·(F

_{y}·t /m)²/2 − m·V_{x}²/2 == F

_{x}·V_{x}·t + F_{x}²·t²/(2m) ++ F

_{y}²·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.

certain period of time equals to an increment of a kinetic energy of

this object

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