Tuesday, January 22, 2019

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 /

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 /

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 / 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]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]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.

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
.

Thursday, December 27, 2018

Unizor - Physics4Teens - Mechanics - Problems on Power





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



Problems on Mechanical Power



Problem A



A car engine accelerates a car of mass m from the state at rest to speed V during the time T with constant acceleration along a straight line.

Ignore loss of mass due to burning fuel.

(a) What is the power of the car engine as a function of time?

(b) If engine's power is proportional to amount of fuel supplied to it
in a unit of time (fuel burring speed), how the fuel burning speed will
change over time to assure the kind of motion described in this problem?


Solution:



(a) a = V/T

F = m·a = m·V/T

S(t) = a·t²/2 = (V/T)·t²/2

W(t) = F·S(t) = (m·V/T)·(V/T)·t²/2 = m·V²·t²/(2T²)

P(t) = dW(t)/dt = (m·V²/T²)·t



(b) The fuel burning speed is linearly increasing with time. The faster a
car goes, as it accelerates, - the more power should be supplied to
assure the constant acceleration, and the faster fuel is burning.





Problem B



A car engine supplies constant power P to a car of mass m.

Find its speed and acceleration as a function of time.

Is speed linearly increasing with time?



Solution



We will use the general formula for power as a function of mass, speed and acceleration:

P(t) = m·V(t)·a(t)

or, since a(t)=dV(t)/dt,

P(t) = m·V(t)·dV(t)/dt

Therefore,

P(t)·dt = m·V(t)·dV(t)

In our case P(t) is constant P. Therefore,

dt = m·V(t)·dV(t)

Integrating this by time from t=0 to t,

P·t = m·V²(t)/2

From this we can find speed V(t):

V(t) = √2P·t/m

Acceleration is

a(t) = dV(t)/dt = √P·/(2m·t)

Speed is not linearly increasing with time, it is proportional to a
square root of time, which means that acceleration is monotonically
diminishes, while speed increases.





Problem C



A car engine supplies constant power P to a car of mass m.

How long would it take for a car to reach speed Vmax?



Solution



As we know,

P(t) = m·V(t)·a(t)

Since power P(t)=P - is constant, and acceleration is the first derivative of speed by time, this can be written as

P = m·V(t)·dV(t)/dt

Therefore,

dt = m·V(t)·dV(t)

Integrating this by time from t=0 to t,

P·t = m·V²(t)/2

From this we can find time t as a function of speed:

t = m·V²(t)/(2P)

Therefore, the time at the moment the speed is equal to Vmax equals to

tmax = m·V²max /(2P)

Obviously, the more powerful an engine is - the shorter will be the time interval it takes to achieve needed speed.



As an example, consider a car 2012 Tesla Model S 85 kWh .

It has a mass of 2108 kg and its engine develops the power of 310 Kw.

According to the formula above, the time it takes for this car to reach the speed of 60 miles/hour (26.8 m/sec) equal to

2108·26.8²/(2·310000) = 2.44(sec)

So, in theory, this car is capable to reach the speed of 60 miles/hour in just under 2.5 sec.

Wednesday, December 26, 2018

Unizor - Physics4Teens - Mechanics - Power





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



Definition of Mechanical Power



To analyze the motion, we often use a concept of speed.

Let's assume that an object moves in some inertial reference frame, and
the distance covered by it from its initial position along its
trajectory is a function of time S(t).

Recall the definition of speed of an object as the rate, at which this object covers the distance along its motion along a trajectory.

In case of a uniform motion we can simply divide the distance S, covered during time t, by the time t to get the speed:

V = S/t

In case of non-uniform motion the speed changes and at any particular moment of time t an instantaneous speed can be calculated using differentials:

V(t) = dS(t)/dt



To analyze the mechanical work performed to achieve certain results, we often use a concept of power.

Let's assume that something or someone performs certain work and, as the time goes by, the work performed is a function of time W(t).

The power is the rate, at which the work is performed.

If during the time t the work performed is W, we define the average power of the whoever or whatever performs the work as

P = W/t

Most likely, at equal in length but different time intervals the amount
of work performed will be different. For example, when a car starts, its
engine should give a car an acceleration, which requires more work per
unit of time than to maintain a constant speed on a smooth straight
road.

In cases like this we can talk about an instantaneous power as a function of time that can be calculated using differentials:

P(t) = dW(t)/dt



Consider an example of an object in uniform motion against the force of friction with a constant speed V.

The force F that moves it forward must be equal in
magnitude and opposite in direction to the force of friction to maintain
the constant speed. Since the friction is constant, the force F must be constant as well.

The distance S covered as a function of time t is

S(t) = V·t

Therefore, the work performed by the force F during the time t is

W(t) = F·S(t) = F·V·t

From the definition of power follows that the power this force F exhorts is

P(t) = dW(t)/dt = F·V

As an example, the car engine exhorts the same power and consumes the
same amount of gas per unit of time, if the car uniformly moves along a
straight road. This power is used to generate a force sufficient to
overcome the friction of wheels and air resistance.



Consider a more general case, when the motion is not uniform.

Assume, an object of mass m moves as a result of action of force F(t), where t is time. The distance it covers is S(t).

Then during an infinitesimal time interval from t to t+dt the work performed by this force will be

dW(t) = F(t)·dS(t)

Considering the Newton's second law,

F(t) = m·a(t),

where a(t) is acceleration.

Increment of distance is

dS(t) = V(t)·dt,

where V(t) is an object's speed.

Also, by definition of acceleration,

a(t) = dV(t)/dt

Therefore,

dW(t) = (m·dV(t)/dt)·V(t)·dt =

= m·V·
dV(t)


Power exhausted by this force is, therefore

P(t) = dW(t)/dt =

= m·V·
dV(t)/dt =

= m·V(t)·a(t)




From the definition of power as amount of work per unit of time or, more precisely, the first derivative of work by time

P(t) = dW(t)/dt

follows that the unit of measurement of power is joule/sec called watt.

Expanding the definition of joule as newton·meter,

1 watt = 1J/sec = 1N·m/sec

Obvious extensions of unit of power watt are

kilowatt = 1,000 watt and

megawatt = 1,000,000 watt.



There is an old unit of power called horsepower.

Metric horsepower, derived from lifting up against a force of gravity on Earth a weight of mass 75 kg with a constant speed of 1 m/sec, is related to watt unit as

1(metric HP) =

=75(kg)·9.8(m/sec²)·1(m/sec)≅

≅ 735.5(W)


For historical reasons there is also a mechanical horsepower, defined as 33,000 pound-feet per minute, related to watt unit as

1(mechanical HP) ≅ 745.7(W)



So, a car engine of 200 mechanic horsepower has the power of about 149,140 watt.



Watt, as a unit of measurement, was called in honor of James
Watt, an 18th century Scottish scientist who was one of the first to
research a concept of power, developed steam engines and measured the
power of a horse.



Now let's address the concept of power in a case of rotation with constant angular speed. An example is lifting a bucket of water from a well.

Assume, a bucket of water has a mass m and we lift it with constant linear speed V with an angular speed of the well's wheel ω=V/R, where R - radius of a well's wheel.



Since the speed is constant, the force F that acts on a bucket equals to m·g, where g - acceleration of the free fall.

At the same time, if the wheel is turned by some motor and R is the radius of its shaft, the motor manufacturer provides technical characteristic not only of the power, but also of a torque of a motor.

Remember that the torque equals to

τ = F·R



So, on one hand, we have expressed the power of a motor P in terms of unknown force F and linear speed of a bucket:

P = F·V = F·R·ω

On another hand, we expressed the torque of this motor in terms of the same unknown force F and a radius of its shaft:

τ = F·R

Substituting torque τ for F·R in the formula for power, we can find the relationship between the power of a motor and is torque:

P = τ·ω



Notice the similarity between the formula for power in case of uniform motion along a straight line P=F·V and formula for power in case of rotation with constant angular speed P=τ·ω. Instead of force F in case of straight line motion, we use torque τ for rotation and, instead of linear speed V for straight line motion, we use angular speed ω for rotation.



Let's check this with real data about a particular engine.

Below is a graph representing the power and torque of Ford Motor Company 6.7L Power Stroke diesel V-8.



As you see, the power and torque grow relatively monotonically until
some engine limitations start playing significant role. While in the
area of monotonic growth, we can take a particular angular speed, say,
1400 RPM (revolutions per minute) and see that the power of an engine
equals approximately 174 HP (mechanical horsepower) and the torque is
about 650 Lb-Ft (pound-feet).

Let's check if the relationship between power and torque derived above is held in this case.

First of all, we transform all units into standard physical measures defined in SI:

1 RPM (revolutions per minutes) =

= 2π radian per minute =

= 2π/60 radians/sec =

= 0.1048 rad/sec


1 HP = 745.7 W = 745.7 J/sec

1 Lb-Ft = 1.35582 J

Substituting all the above, we see that

angular speed is equal to

1400·0.1048 = 146.6 rad/sec

power is

174·745.7 = 129,751.8 J/sec

torque equals to

650·1.35582 = 881.3 J



Now we can check the relationship between the power, torque and angular speed

P = τ·ω

Indeed,

881.3·146.6 = 129,198.6

As we see, the difference between power and a product of torque by
angular speed is minimal, attributable to imprecise measurement of the
parameters and graph reading.

Tuesday, December 11, 2018

Unizor - Physics4Teens - Mechanics - Work - Problems











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



Problems on Mechanical Work



Problem A



An object of mass m is pushed up a plane inclined to horizon at angle φ with constant acceleration a to a height h against a friction with coefficient of kinetic friction μ.

(a) What is the amount of work necessary to achieve this goal?



(b) Consider the following real conditions of this experiment:

m = 10kg(kilograms)

g = 9.8m/sec²(meters per sec²)

h = 0.5m(meters)

φ = 30°(degrees)

μ = 0.2

a = 0.1m/sec²(meters per sec²)

The work W=F·S is measured in N·m=kg·m²/sec² units, called joule and abbreviated as J.

Calculate the work W performed in this experiments in joules.



Solution

Let F be the force that pushes the object up the slope, P=m·g is the object's weight, R is the force of resistance from the friction, S is the distance this object moves to reach the height h.

(a) F−P·sin(φ)−R = m·a

R = P·cos(φ)·μ

W = F·S

S = h/sin(φ)

F = P·sin(φ) + R + m·a =

= m·[g·sin(φ) + g·cos(φ)·μ + a]


W = F·h/sin(φ) =

=m·h[g+g·cot(φ)·μ+a/sin(φ)]


(b)
W = 67J(joules)





Problem B



An object of mass m is pushed by a constant force F down a slope of a plane inclined to horizon at angle φ. Initially, it's at rest, the final speed is V. There is a friction with coefficient of kinetic friction μ.

(a) What is the time t from the beginning to the end of the object's motion?

(b) What is the distance S this object moved until reaching the final speed V?

(c) What is the amount of work W performed by this force?

(d) Consider the following real conditions of this experiment:

F = 20N(newtons)

m = 10kg(kilograms)

g = 9.8m/sec²(meters per sec²)

V = 0.5m/sec(meters per sec)

φ = 5°(degrees)

μ = 0.7

Calculate the work W performed in this experiments in joules, distance S and time t of motion.



Solution

P=m·g is the object's weight,

R = P·cos(φ)·μ is the force of resistance from the friction,

S is the distance this object moves to reach the speed V,

a is the acceleration of this object.

Then from the Newton's Second Law

F+P·sin(φ)−R = m·a

(we added P·sin(φ) because an object moves down a slope)

Now we can find an acceleration of our object:

F+m·g·sin(φ)−m·g·cos(φ)·μ =

= m·a


a = F/m + g·sin(φ) − g·cos(φ)·μ

Knowing acceleration a and the correspondence between acceleration, final speed and time V=a·t, we can determine time:

(a) t = V/a

Now we can find the distance

(b) S = a·t²/2 = V·t/2

The work performed by the force F equals to

(c) W = F·S

(d) a = 0.066(m/sec²)

t = 7.576(sec)

S = 1.894(m)

W = 37.879(joules))





Problem C



An object of mass m is dropped down from a certain height above a surface of some planet. At the moment it hits the ground its speed is V.

What is the work W performed by the force of gravity?



Solution

Let the acceleration of free falling on this planet is a, the time of falling t and the height S.

Then

F = m·a

V = a·t

t = V/a

S = a·t²/2 = V²/(2·a)

W = F·S = m·V²/2

Notice, the work depends only on mass and final speed - the results of
action, not on the height, nor acceleration of free falling, nor on
time.

Friday, October 12, 2018

Unizor - Physics4Teens - Mechanics - Work - Golden Rule of Mechanics





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



Golden Rule of Mechanics



Let's summarize what we have learned about mechanical work in the previous lecture.



1. In a simple case of motion along a straight line with a constant force F acting along a trajectory, the most important parameter that quantifies the result of the action is work defined as W=F·S, that fully characterizes and is fully characterized by speed V of an object:

W = F·S = m·V²/2

In particular, it means that increasing the force by a factor of N and decreasing the distance it acts by the same factor of N would result in the same final speed of an object.

So, we can "win" in distance, but we will "lose" in force and vice versa.

This is the first example of the Golden Rule of Mechanics - there
are many ways to achieve the result, you can reduce your distance, but
you will have to increase the force or you can reduce the force, but you
will have to increase the distance.

In short, as we mentioned above, whatever you win in distance you lose in force and vice versa.



2. In case of a constant force acting at an angle to a straight line trajectory, the difference is only a factor cos(φ), where φ is an angle between a force and a direction of a trajectory. In vector form it represents the scalar product (F·S).

So, the Golden Rule works exactly as above.



3. Recall the formula for work of a force F pushing an object of weight P up along an inclined plane of the length S making angle φ with horizon to the height H:

W = F·S = P·H

We've proven this in the previous lecture and, as you see, it's
independent of the slope of an inclined plane. However, the minimum
effort we have to apply as a force to move an object up the slope is F=P·sin(φ), while the distance equals to S=H/sin(φ).

We can reduce the effort (the force F) by using an incline of a smaller slope, but that would lengthen the distance S we have to push an object.

So, again, we see the Golden Rule in action.



4. Consider lifting some heavy object of the weight P using a lever, applying the force F to the opposite to an object end of the lever.



This is a case of equilibrium in rotational motion and the balance can
be achieved by equating the moments of two forces acting against each
other:

F·Lf = P·Lp

If Sf is the distance the force F acts down and Sp is the distance our object moves up,

Sf/Sp = Lf/Lp and

F·Sf = P·Sp

By using a lever with longer arm Lf for the application of force, we can proportionally reduce the force F
achieving the same result - lifting an object to certain height. An
inverse is true as well - we can shorten the arm and proportionally
increase the force.

In any case, the Golden Rule of Mechanics is observed: "winning"
in force - proportionally "losing" in distance or "winning" in distance"
- proportionally "losing" in force.



All the above examples emphasize the importance of the concept of mechanical work as the quantitative measure and characteristic of the purpose and the result of applying a force. The so-called Golden Rule of Mechanics is just a catchy term that underscores the importance of the concept of work.



Acting with the force to achieve certain goal necessitates performing
certain amount of work that depends on the goal, not on how we achieve
it
.