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.