Unizor - Physics4Teens - Energy - Potential Energy - Gravity

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

Energy of Gravity

An object of mass m is raised on certain height h above the surface of the Earth.
Analyze its potential energy, as a function of height h.
Do not assume that the force of gravity is constant, but rather use the Newton's Law of Universal Gravitation.
Assume that the radius of Earth is R and its mass is M.

Solution:

If x is the distance from the center of the Earth to an object, the force of gravity equals to
F(x) = G·M·m/x²
(where G is a gravitational constant approximately equaled to 6.674·10^-11N·m²·kg⁻²).

Potential energy of an object at height h above the ground is the work performed by the force of gravity as an object falls down to the surface of the Earth from the distance R+h from the center to the distance R.

Since the force of gravity varies, we should use the calculus to calculate this work.
An infinitesimal increment of work dW(x), assuming the force F(x) acts at a distance dx, equals to
dW(x) = F(x)·dx

Integrating this by x from R+hto R, we get the full work and the potential energy of an object at height h above the ground:
E_pot = W(h) = ∫^R_R+hF(x)·dx =
= G·M·m·[1/R − 1/(R+h)]

Obviously, when an object is at the ground level (h=0), its potential energy equals to zero and, as it moves higher (h is increasing), its potential energy grows to its maximum value of G·M·m/R, as the object moves farther and farther from the ground.

When the height of the object his significantly smaller than the radius of the Earth (h << R), our formula for potential energy can be approximated with a simpler expression:
E_pot = W(h) =
= G·M·m·[1/R − 1/(R+h)] =
= G·M·m·h/[R·(R+h)] ≅
≅ G·M·m·h/R² = m·g·h,
where g=G·M/R² is the free fall acceleration at the ground level, approximately equaled to
g ≅ 9.81 m/sec²
and P = m·g is the weight of an object at the ground level, assumed constant for small heights h.
In this case the potential energy is simply a product of weight (force of gravity) and distance - a classical expression for work performed by a constant force:
E_pot = W(h) ≅ P·h

Let's also analyze the kinetic energy of the object.
Firstly, we assume that our object starts its free falling from the height H above the ground with zero initial velocity.

Its potential energy at any height h we know from the above calculation:
E_pot = G·M·m·[1/R − 1/(R+h)]
Its kinetic energy depends on its speed V:
E_kin = m·V²/2
The problem is, we don't know how the speed depends on the height h.

What we do know is the dependency of the force of gravity F on the height h:
F = G·M·m/(R+h)²
We also know that speed V is the first derivative of height hby time t and the acceleration is the second derivative:
V(t) = h'(t)
a(t) = h"(t)
Therefore, according to the Newton's Second Law,
F = m·a = G·M·m/(R+h)²
Now we have an expression for acceleration a:
a = G·M/(R+h)²
Let's underscore that acceleration is the second derivative of distance h, as a function of time t:
h"(t) = G·M/[R+h(t)]²

We will not attempt to solve this differential equation to get a function h(t), take its derivative to get the speed and substitute into a formula for kinetic energy.
Instead, we will apply a clever trick.
Let's multiply the last expression by 2h'(t):
2h'(t)h"(t) =
= 2G·M·h'(t)/[R+h(t)]²
The reason for this operation is the following.
The left side of this equation is the derivative of the square of the speed:
d/dt{[h'(t)]²} = 2h'(t)·h"(t)
and on the right side of this equation we find another derivative:d/dt{2G·M/[R+h(t)]} =
= 2G·M·h'(t)/[R+h(t)]²
Therefore, we have the equality of the derivatives:
d/dt{[h'(t)]²} =
= d/dt{2G·M/[R+h(t)]}

If derivatives are equal, the functions differ by a constant:
[h'(t)]² = 2G·M/[R+h(t)] + C
The constant C can be determined, using the conditions at the beginning of motion at time t=0:
h(0) = H and h'(0) = 0
Therefore,
[h'(0)]² = 2G·M/[R+h(0)] + C
[0]² = 2G·M/[R+H] + C
C = − 2G·M/[R+H]
Now we can write
[h'(t)]² =
= 2G·M/[R+h(t)]−2G·M/[R+H]
From this we can express the kinetic energy
E_kin = m·V²(t)/2 =
= m·[h'(t)]²(t)/2 =
= G·M·{1/[R+h(t)]−1/[R+H]}

Since our task was to express the energy in terms of height above the ground, we can simply write it as
E_kin = G·M·[1/(R+h)−1/(R+H)]

Finally, let's find the full mechanical energy of an object falling from the sky.
E_pot = G·M·m·[1/R−1/(R+h)]
E_kin = G·M·[1/(R+h)−1/(R+H)]
E_full = E_pot + E_kin =
= G·M·[1/R−1/(R+H)]
As we see, the full mechanical energy is constant, it does not depend on the height. As an object falls from the sky, its potential energy is decreasing, but kinetic energy increasing, and the sum of both types of energy remains constant.

Unizor - Physics4Teens - Energy - Potential Energy of a Spring

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

Energy of Spring

The energy discussed in the previous introductory lecture was an energy related to constant force acting on an object. In this lecture we will address a more difficult case of a variable force of a spring acting horizontally on an object according to the Hooke's Law.
Let's remind the Hooke's Law. It states that the force of a spring, fixed on one end, onto a point-object attached to its free end is proportional to the length the spring is extended or compressed from its neutral position.
The corresponding formula for the force F of a spring onto an object is
F = −k·L
where L is a displacement of the spring's free end and k is a coefficient of elasticity - a specific characteristic of "stiffness" of each spring.
The minus sign signifies that the direction of the force is opposite to a displacement. The frame of reference is associated with the neutral position of a spring and the X-axis is directed towards spring's stretching. Then, for positive displacement L(stretching) the force is directed towards neutral point (that is, negative), while for negative displacement (compression) the force also directed towards a neutral position, (that is, positive).

At this point we recommend to refresh the material in the "Work and Elasticity" lecture of the "Mechanical Work" chapter of the "Mechanics" part of this course.
As stated in that lecture, the total amount of work W to compress (or stretch) a spring with elasticity k by length Lequals
W = k·L²/2
regardless of how exactly we compress or stretch a spring.

Assume now that we have compressed a spring by the length L, performing work equaled to W=k·L²/2. At a spring's end we have attached an object of mass m. Then we let the spring go by itself with the object attached to its end.

Intuitively, we think that, at first, the spring should return back to its neutral position. During this time the object's speed will increase because the force of elasticity of a spring will push it.

But then, when a spring reaches its neutral position, the object, having certain speed, will continue its move by inertia, stretching the spring. During this time the force of elasticity of a spring will slow down the movement of the object until it stops at the end of the stretching process.

At that time the force of elasticity will pull the object back to a neutral position, increasing its speed.

At the neutral position object continues moving by inertia, compressing a string, until the force of elasticity stops its movement when the spring will be in the compressed position, as in the beginning of our process.

At this point the process repeats itself and can continue like this indefinitely.

Well, our intuition is correct, a spring with an object attached to it will oscillate indefinitely. Let's analyze quantitatively these oscillations.

Our first goal is to find the dependency of the object's position on time. This can be done using the Newton's Second Law, taking into account the initial position of the object at the end of a compressed spring with no initial speed.
Let S(t) be a function describing the position of the object at the end of a spring relatively to its neutral position with stretching being a positive direction. Then S(0)=−L and S'(0)=0. The acceleration is the second derivative of S(t), that is a=S"(t). The force of elasticity is proportional to S(t) as F=−k·S(t). Now the Newton's Second Law F=m·a looks like this differential equation:
−k·S(t) = m·S"(d)
with two initial conditions:
S(0) = −L
S'(0) = 0
This differential equation is fully solvable and its general solution is
x(t) = C₁·cos(t·√k/m) +
+ C₂·sin(t·√k/m)

Constants C₁ and C₂ can be determined using the initial conditions:
C₁ = −L
C₂ = 0
Therefore,
S(t) = −L·cos(t·√k/m)

This function describes simple sinusoidal oscillations with amplitude L and period
T=2π√m/k.

We can also calculate the speed of an object as the first derivative of distance:
V(t) = S'(t) = dS(t)/dt =
= L·√k/m·sin(t·√k/m)

During the first quarter of this time an object accelerates under the influence of the elasticity of a spring, moving from the extremely compressed position on a spring to its neutral position. During the next quarter of the period the spring's elasticity slows it down, while it moves to extremely stretched position of a spring. During the next quarter of the period elasticity of a spring accelerates it again to a string's neutral position. Finally, the elasticity slows the object down, as it moves to the initial most compressed position.

Let's calculate the work that spring performs during the first quarter of its period from complete compression to a neutral point by accelerating the object from initial speed V(0)=0to its maximum at a spring's neutral position V(T/4).
It can be done by two methods.

1. By integrating by displacement of a spring's end from value −L (full compression) to value 0 (neutral position)
W_[−L,0] =
= ∫⁰_−LF(S)·dS =
= ∫⁰_−L(−k·S)·dS =
= −k·S²/2|⁰_−L = k·L²/2

2. By integrating by time from t=0 to t=T/4=(π/2)·√m/k(calculations are more complex but lead to the same result)
W_[0,T/4] =
= ∫₀^T/4F(S(t))·dS(t) =
= ∫₀^T/4F(S(t))·(dS(t)/dt)·dt =
= ∫₀^T/4F(S(t))·V(t)·dt =
= ∫₀^T/4k·L·cos(t·√k/m)·
·L·√k/m·sin(t·√k/m)·dt =
= ∫₀^T/4k·L²·cos(t·√k/m)·
·sin(t·√k/m)·d(t·√k/m) =
= ∫₀^T/4k·L²·sin(t·√k/m)·
·d(sin(t·√k/m)) =
= k·L²sin²((1/2)π√m/k·√k/m)/2
After cancellation of factors and taking into account that sin(π/2)=1 we obtain the same expression for work:
W_[0,T/4] = k·L²/2

What we have proven now is that the same amount of work done to compress a spring (W=k·L²/2) is performed by a spring when it returns to its neutral position.

That's why we say that the spring accumulates potential energy, when it's compressed, and releases it, when it returns back to neutral position.

Now, let's see where this energy, released by a spring during its return to a neutral position, goes. Our assumption is that it is transformed into kinetic energy of the object attached to its free end. Let's check it out.
E_kin = m·V²(T/4)/2 =
= m·L²·(k/m)·sin²(π/2)/2 =
= k·L²/2
So, kinetic energy of the object, when a spring is in the neutral position, exactly equals to a spring's potential energy when it's fully compressed.
Compressing a spring results in transferring some external energy to its potential energy. Then, as the spring returns to its neutral position, it releases this potential energy into a kinetic energy of the object attached to its end. At the neutral position the potential energy of a spring is zero, but kinetic energy of the object equals to potential energy of a spring in a compressed position.

Our last step is to calculate full energy of the system (potential energy of a spring and kinetic energy of the object attached to it) at any moment of time t.
E_pot(t) = k·S²(t)/2 =
= k·L²·cos²(t·√k/m)/2
E_kin = m·V²(t)/2 =
= m·L²·k/m·sin²(t·√k/m)/2
E_full = E_pot + E_kin = k·L²/2
(since sin²(φ)+cos²(φ)=1)

As we see, the full energyremains constant, it does not depend on the attached object's mass, only on the properties of the spring - coefficient of elasticity k and its initial displacement from the neutral position.

Unizor - Physics4Teens - Energy - Potential Energy - Introduction

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

Potential Energy - Introduction

Potential energy is a quantitative characteristic of an object's position relative to other objects or environment that are a source of some forces. These forces would result in motion of the object under consideration (if no counteracting forces present) and, therefore, would result in performing certain work related to this motion.

Typical example of potential energy is related to a position of some object in the gravitational field.
Since a planet Earth attracts all objects, an object raised to a certain height above the ground will fall, if no support is provided. The motion of falling object is the result of work of the forces of gravity. Therefore, by definition, an object raised above the ground has certain amount of potential energyequal to work performed by the forces of gravity to move it down to the ground.

In a simplified case of a small object of mass m, raised on small (relatively to the size of the Earth) height h, the force of gravity, acting on this object, is constant and is equal to the object's weight P = m·g, where g ≅ 9.8m/sec² is the acceleration of the free fall near the Earth's surface.
If no support is provided, the object will fall straight down. Thus, the constant force of gravity P would perform work, pushing the object down at a distance h, thus performing work
W = P·h = m·g·h

Let's examine the relationship between an object's potential energy and its kinetic energy as it falls down from certain initial height.
Assume an inertial frame of reference associated with Earth and X-coordinate directed straight up perpendicularly to the ground with ground level be at coordinate X=0.
Initial position of the object is, therefore, at X-coordinate equal to X=h. The falling object moves with constant acceleration −g (negative, since it moves from positive X=h to X=0). Initial speed of the object is zero.
Therefore, the time-dependent motion X(t) can be described by the kinematic equation
X(t) = h − g·t²/2
The object's velocity is directed vertically down and, as a function of time, is the first derivative of the X(t):
V(t) = X'(t) = −g·t
(which is obvious from the fact that the motion is with constant acceleration −g and zero initial speed)

As object falls down, its height diminishes, which means that its potential energy also diminishes. At any given moment of time t this energy equals to
E_pot(t) = m·g·X(t) = m·g·(h−g·t²/2)

At the same time the kinetic energy of the object is increasing, since its speed V(t)is increasing. At any given moment of time t this energy equals to
E_kin(t) = m·V²(t)/2 = m·g²·t²/2

Remarkably, the full mechanical energy of this object, which is a sum of its potential energy and kinetic energy is constant:
E_full = E_pot + E_kin = m·g·h,
which is its potential energy in the beginning of motion (when its initial speed and, therefore, kinetic energy are zero).

The total time in motion from the initial position above the ground at height h to the ground can be obtained by resolving the equation X(t)=0 for variable t:
X(t) = h − g·t²/2 = 0
t_fall = √2h/g
At that time of touching the ground the potential energy of an object equals to zero and its kinetic energy equals to
E_kin(t_fall) = m·V²(t_fall)/2 = m·g²t²_fall/2 = m·g·h

In the beginning the potential energy is at its maximum and kinetic energy at zero. At the end the potential energy is at zero, while kinetic energy at its maximum, but their sum is always the same. So, it appears that, as an object falls, its potential energy is transformed into kinetic energy, preserving their sum - the full mechanical energy.

Unizor - Physics4Teens - Energy - Problems on Kinetic Energy

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

Problems on Kinetic Energy

Problem A

A car engine accelerates a car of mass m from the state at rest to some maximum speed during the time T with constant acceleration a along a straight line.
Ignore loss of mass due to burning fuel.
What is the kinetic energy E_kinof a car at the end of this period of acceleration and what is the work W performed by the car engine during this time?

Solution:

V_end = a·T
E_kin = m·V²_end /2 = m·a²·T²/2

W = F·S = (m·a) · (a·T²/2) =
= m·a²·T²/2
W = E_kin = m·a²·T²/2

Problem B

A driver of a car, going with speed V, sees an obstacle and strongly presses on breaks, so its wheels stop spinning. The car slows down to complete stop by the force of friction only at a distance S from the spot when it started to break.
What is the coefficient of kinetic friction μ of car's wheels against the ground?
Assume, the free fall acceleration is g.

Solution

Let m be an unknown mass of the car.
Let W be a work of the force of friction to stop the car.
Let E_kin be a kinetic energy of the car in the beginning of breaking. The ending kinetic energy is zero, since the speed at the end is zero. So, the initial kinetic energy should be equal to the work required to stop the car.
Let F be a constant force of friction equal to the car's weight multiplied by a coefficient of kinetic friction.
E_kin = m·V²/2 = W
W = F_·S
F = W/S = m·g·μ
μ = W/(m·g·S) = m·V²/(2·m·g·S)
μ = V²/(2·g·S)
Notice independence of the car's mass.


Problem C

A point-object A of mass m is freely rotating on a thread of a length L that forms an angle φwith vertical.
What is its kinetic energy?

Solution

T is a force of tension of the thread. We can represent it as a sum of two components - T_vthat balances the weight P and T_h that keeps an object on a circular trajectory (centripetal force).
T_v = m·g
T_h = T_v·tan(φ) = m·g·tan(φ)
This centripetal force causes centripetal acceleration
a = V²/R
(see a chapter Mechanics - Superposition of Forces - Resultant Forces - Example 2 in this course)
where V is a linear speed of an object and R is a radius of a circular trajectory, which is equal to L·sin(φ).
Therefore,
T_h = m·a = m·V²/R
Hence,
m·g·tan(φ) = m·V²/R
Kinetic energy is
E = m·V²/2 =
= m·g·tan(φ)·R/2 =
= m·g·L·tan(φ)·sin(φ)/2


Problem D

An ideal pendulum of mass mwith a thread length L performs small harmonic oscillations with its angle of deviation from the vertical φ conforming to the equation
φ(t) = φ₀·cos(√g/L ·t)
where φ₀ is the initial deviation from the vertical, g - free fall acceleration and t - time.
What is its kinetic energy at the lowest point of its trajectory?

Solution:

We recommend to refresh the properties of a pendulum in the "Mechanics" part of this course in the chapter "Pendulum, Spring".
Let's find the period of oscillations.
Since function y=cos(x) has a period 2π, function y=cos(k·x)has period 2π/k.
In our case the period of oscillation T is equal to
T = 2π·√L/g 
The angular velocity is a derivative of the angle of deviation φ(t):
ω(t) = dφ(t)/dt =
= −φ₀·√g/L ·sin(√g/L ·t)
At the lowest point of a trajectory the time equals to 1/4 of a period.
Therefore, we can calculate the angular velocity at this point by substituting
t = T/4 = (1/4)·2π·√L/g =
= (π/2)·√L/g 
which gives the angular speed at this point as
ω(T/4) =
= −φ₀·√g/L ·sin(√g/L ·T/4) =
= −φ₀·√g/L ·
·sin(√g/L ·(π/2)·√L/g ) =
= −φ₀·√g/L 
The linear speed of a rotating object equals to its angular speed, multiplied by a radius.
Therefore, linear speed V at the lowest point equals to
V = L·ω(T/4) =
= −φ₀·√g·L 
The kinetic energy at this point is
E = m·V²/2 = m·g·L·φ₀²/2
IMPORTANT NOTE:
The following equation for small harmonic oscillation used in this problem
φ(t) = φ₀·cos(√g/L ·t)
was obtained in "Mechanics" part of this course in chapter "Pendulum, Spring" by simplification of the differential equation that described the motion of the pendulum. In particular, it was assumed that for small oscillations the angle of deviation from the vertical φand its sine sin(φ) are approximately equal. The exact value of the kinetic energy of a pendulum at its lowest point can be derived using its potential energy at the top position, which we will address in the next chapter, and which is equal to 2m·g·L·sin²(φ₀/2). If in this expression we replace sin(φ₀)with φ₀ (which is acceptable for small angles), we will get the formula for kinetic energy we obtained above.

Unizor - Physics4Teens - Mechanics - Work and Gravity

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



Work and Gravity



Let's consider a force to lift an object above the planet against the
force of gravity. In simple cases, when the height is relatively small
compared with the size of a planet, the force of gravity is considered
to be constant. In this case we will assume that we lift to substantial
height and have to take into consideration the Law of Gravity that tells
that the force of gravity is proportional to masses of objects involved
(a planet of mass M and an object of mass m that we lift) and inversely proportional to a distance r between the objects:

F = G·M·m /r²

where G - a gravitational constant.



Our task is to find the work W needed to lift an object of mass m from a surface of a planet of mass M and radius R to height H above its surface.



The easiest approach is to represent all parameters as functions of the distance r between a center of a planet and an object.

Then the force of gravity as a function of r is

F(r) = G·M·m /r²

An infinitesimal increment (differential) of work equals

dW(r) = F(r)·dr =

= G·(M·m /r²)·dr



Let's integrate the differential of work on an interval of r [R,R+H]:

∫_[R,R+H]dW(r) =

= G·M·m·∫_[R,R+H](1/r²)dr =

= G·M·m·[1/R − 1/(R+H)] =

= G·M·m·H / [R·(R+H)]



For small (relatively to radius of a planet R) height H the expression in curly brackets above is approximately equal to

G·M·m /R² = m·g

where g = G·M /R² is an acceleration of the free falling on a planet's surface.

By definition, m·g is the weight P of an object on a surface of a planet, which we consider a constant force in this approximation.

So, our formula for work for small height above the planet is reduced to simple expression

W = P·H

which is a base "force times distance" expression that defines the work
in simple case of constant force acting along a trajectory.



For large height H we cannot ignore the change of gravity
as an object moves far from the planet, and the exact formula must be
used to calculate the work.



As in other cases, the work depends only on characteristics of
interacting objects (their masses in this case) and the result of work
(lifting on certain height), not the way how we achieve this result.

Unizor - Physics4Teens - Mechanics - Work and Elasticity

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



Work and Elasticity



Let's consider a slightly more complicated case of a variable force.

A good example of this is the work performed by a force to stretch or
compress a spring, in which case a force linearly depends on a length a
spring is stretched or compressed.



Consider a spring with elasticity coefficient k that force F compresses by length S.

Consider further that a spring is compressed by the same length in equal
time interval, that is its end, where the force is applied, is
uniformly moves under the force of compression with constant speed V.

Our task is to find the amount of work needed to compress a spring by the length L.



In this case we can calculate the length of compression as a function of time

S(t) = V·t

According to the Hook's Law, the force of compression is proportional to the length of compression:

F = k·S

Since the force and the length of compression are time-dependent, this can be written as

F(t) = k·S(t) = k·V·t



An infinitesimal increment (differential) of work equals

dW(t) = F(t)·dS(t) = F(t)·V·dt =

= k·V·t·V·dt = k·V²·t·dt



The time interval T needed to compress a spring by length L with speed of compression V equals to T = L/V

Let's integrate the differential of work on a time interval [0,T]:

∫_[0,T]dW(t) = k·V²·∫_[0,T]t·dt =

= k·V²·T²/2 = k·L²/2



Remarkably, the work does not depend on speed V of compression, only on the length L the spring is compressed and its elasticity k.



We can easily generalize this by getting rid of dependency on the speed of compression in our calculations.

Instead of all parameters being functions of time t, let's use the length of compression S as a base variable.

Since F(S) = k·S,

dW(S) = F(S)·dS = k·S·dS

We can integrate it on an interval S∈[0,L] getting

∫_[0,L]dW(S) = k·∫_[0,L]S·dS =

= k·L²/2



As you see, this is a more general (and simpler!) derivation of the same
formula that does not depend at all on how we compress the spring, only
on its elasticity and a length of compression.



So, total amount of work W to compress a spring with elasticity k by length L equals

W = k·L²/2

regardless of how exactly we compress a spring.

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 v_end. What will be the work force F would perform?



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_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 W performed by our force F during the distance S:

W = F·S = m·(v²−v²_end)/2 =

= m·v²/2 − m·v²_end)/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 v_beg to v equals to

W = m·v²/2 − m·v²_beg/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, ... m_N are moving with speeds v₁, v₂, ... v_N then their 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 m initially moves along the X-axis. The X-component of its velocity vector is V_x and its Y-component of a velocity vector V_y is zero. Then its speed V (a scalar) equals to its X-component V_x (a scalar) and its kinetic energy is

E_k = m·V²/2

where V = V_x



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: F_x acting along the X-axis and F_y acting along the Y-axis.

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 /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 V_x and acceleration a_x, the distance covered by our object along the X-axis equals to

S_x = V_x·t + a_x·t² /2

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

S_y = a_y·t² /2



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

S_x = V_x·t + F_x·t² /(2m)

S_y = F_y·t² /(2m)



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

W = F_x·S_x + F_y·S_y =

= F_x·V_x·t + F_x²·t²/(2m) +

+ F_y²·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 /m

The Y-component of the velocity will be equal to

V_y^t = a_y·t = F_y·t /m



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

E_k^t = m·(V^t)² /2

where V^t is the speed at time t.
Using the Pythagorean Theorem, we can represent (V^t)² as

(V^t)² = (V_x^t)² + (V_y^t)².
Therefore, 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 /2



The increment of the kinetic energy is

ΔE = E_x^t + E_y^t − E_k

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·(V_x+F_x·t /m)²/2 +

+ m·(F_y·t /m)²/2 − m·V_x²/2 =

= F_x·V_x·t + F_x²·t²/(2m) +

+ F_y²·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.