The Concept of Energy

Reduced to simplest terms, the universe is nothing more than a series of energy exchanges.  As in much of physics, the concept of energy is easy to understand, but difficult to express and work with.

Notice the simplicity of this definition.  If something has the ability to cause a change in the physical or chemical properties of another object it has energy.  If we restrict our study to mechanical energy, then we consider the changes in the state of motion of an object.

There are different classifications of energy, usually named for the source of the force involved; mechanical, gravitational, electrical, nuclear, etc.  Within each of these areas the energy is further classified as either potential or kinetic energy.

Potential energy - Energy due to an object’s position.
Kinetic energy - Energy due to an object’s state of motion.

The chemical potential energy of a substance depends on the bond structure within the material, the nuclear potential energy of a material depends on the composition if its nuclei, the mechanical potential energy of a wind-up car depends on the state of its internal spring.

The kinetic energy of a gas depends on the motion of its molecules, the kinetic energy of a galaxy depends on the movement of its constituent stars, the mechanical energy of a truck depends on its mass and velocity.

You can’t see energy, which makes it a little tough to comprehend, but you can see, feel, and measure when the energy of an object changes.  It is by these changes that we can get a handle on where the energy is and where it is going.  To find out how the universe works: follow the energy.
 

Work and Power        (Qs and Ps for this section)       (back to the top)

When a force is applied to an object which results in a change in motion or physical state of the object, work is done.  If you detect the motion or physical state of the object changing you know work must be done.

Work is the product of a force and the displacement of the object due to that force.  The catch is that the force and the displacement must be along the same line.  If the force is perpendicular to the motion, no work is done.  If the force is at an angle to the motion then only the component of the force along the motion is used.

EX.  Calculate the work done in each of the following situations:

You can see that the unit for work is the combination of force and displacement (Nm).  This has been redefined as the Joule (J) in honor of Henri Joule.

Power is a measure of the rate at which the energy changes or the work is done.  When something is more powerful either it can do the same amount of work in less time, or it can do more work in the same time.

Power is calculated by dividing the energy change by the time, or the work done by the time.

The unit used for power would be a J/s, but this has been redefined as the watt.

A kilowatt (Kw) is often used for large amounts of power.        1 Kw = 1000 w

EX.  If you lift a 1.5 Kg book from the table to a shelf 0.5 m high in 0.75 s,
              a)  what is the amount of work done, and
              b)  what power is required?

          a)  W = Fd       You have to exert a force on the book equal to its weight to lift it to the shelf.
                                                         F = mg = 1.5 Kg(9.8 m/s²) = 14.7 N
               W = 14.7 N (0.5 m)
                   = 7.35 Ns = 7.35 J

          b)  P = W / T = 7.35 J / 0.75 s = 9.8 J/s = 9.8 w

The power of a device is important.  Engines and motors are rated by their power so that they can be properly matched to a specific task.  If the device does not have the capability of providing the required power either it won’t be able to do the work, or if it can it won’t be fast enough.

For your information, the term “horsepower” is the power unit in the english system.  It actually was defined by the inventor James Watt as the amount of work a standard draft horse could do in a second of time.  One horsepower is the same power as 746 watts.

Potential Energy               (Qs and Ps for this section)       (back to the top)

As stated before, potential energy is the ability an object possesses to cause change based on its position.  By winding a spring it has more ability to do something than when it is unwound.  The spring is not actually doing anything, but it could.  This is the nature of potential energy.

The potential energy we use the most is gravitational potential energy.  This is the energy an object has due to its position from the center of the earth.  If there were no gravitational force, there would be no gravitational potential energy.

Since work causes a change in potential energy, the amount of gravitational potential energy an object has depends on the work done on it to change its vertical position.  If you move an object sideways the grqavitational potential energy of the object does not change since it is the same distance from the center of the earth.

Potential energy only changes when the height of an object changes.  It does not matter the path the object takes to get there.  Just look at the beginning and ending heights to calculate the change in energy.

The formula for calculating the gravitational potential energy of an object is:

Where h is the height above or below the zero reference line.  This reference line can be placed anywhere it is convenient in the problem.  Usually, ground level is chosen as the reference line.

The unit of energy was renamed to make it a shorter unit.  The Kg m²/s² is also a Joule (J) of energy.  Notice that to get Joules of energy the mass must be in kilograms and the height in m.

Since work causes an equal change in energy it makes sense that they would have the same unit of measurement.  If you have nothing better to do you can prove to yourself that a Nm and a Kgm²/s²  are equivalent.

EX.  What is the gravitational potential energy of a 1.6 Kg book on a shelf 2.0 m above the floor?

         PE = m g h  = 1.6 Kg (9.8 m/s² )(2.0 m) = 31.4 J    (relative to the floor)
 

EX.  If a 1.5 Kg book is resting on a table 0.8 m above the floor, what is its potential energy
        relative to

 a) the floor,
 b) the table, and
 c) the ceiling which is 1.8 m above?

        a) in this case the floor would be the zero reference height.
                PE = mgh = 1.5 Kg (9.8 m/s² )(0.8 m) = 11.8 J

        b) in this case the table is the zero reference height.  Since the table is at that height, h = 0
            so PE = 0 J.

        c)  in this case the ceiling is the zero reference height.  Since the book is below the zero line,
             its h is negative.
                PE = mgh = 1.5 Kg (9,8 m/s² )(-1.8 m) = -26.5 J
 

A change in potential energy can be calculated from the change in height the object undergoes.
              DPE = PE_f - PE_o = mgh_f - mgh_o = mg (h_f - h_o)

EX.  What is the change in potential energy of a 30 Kg child on a swing if their highest point of the swing is 2.4 m above the ground and the lowest point of the swing is 0.6 m off the ground?

              DPE = mg (h_f - h_o) = 30 Kg (9.8 m/s² )(2.4 m - 0.6 m) = 529 J
        This would be a gain in potential energy as they go up and a loss of potential energy as they
        come down.

Although potential energy can have both positive and negative values, it is not a vector quantity.  You have to keep track of the gains or losses in energy, but the usual math procedures can be used.  No vector treatment is required.
 

Kinetic Energy                  (Qs and Ps for this section)       (back to the top)

Kinetic energy is determined by the motion of the object.  If an object is moving it will exert a force on anything it interacts with.  This force would do work on the other object.  If an object is not moving it has no ability to run into another object and do work.  This is the nature of kinetic energy.

The formula for calculating the kinetic energy of an object is:

Once again we see a combination of the object’s mass and velocity, similar to momentum.  Momentum was a vector, but kinetic energy is not.  Again, you will not have to use vector procedures to work with kinetic energy.  There will be gains and losses of kinetic energy, but normal arithmetic will calculate the answers.

The kinetic energy depends direstly on the mass of the object, but it also depends on the square of the velocity.  Changing the speed a little changes the kinetic energy alot.  You might notice this in driving a car.  It takes quite a bit more work to be done on your car to make it go from 50 mph to 60 mph than to go from 10 mph to 20 mph.

EX.  What is the kinetic energy of a 800Kg car moving at 15 m/s ?

          KE = 1/2 m V² = 1/2 (800 Kg)(15 m/s)²  = 90,000 Kg m²/s² = 90,000 J
 

A change in kinetic energy is calculated like any other change quantity; final energy minus initial energy.

         DKE = KE_f - KE_o = 1/2mV_f² - 1/2mV_o²  = 1/2m(V_f² -V_o² )

EX.  What is the change in kinetic energy for a 1000 Kg car going from 30 m/s to 12 m/s as it enters a speed zone?

         DKE = 1/2m(V_f² -V_o² ) = 1/2(1000 Kg)(12² - 30² ) = -378,000 J

    Notice that the quantity (V_f² -V_o² ) is not the same as (V_f - V_o)² .  Be careful.
 

Changes in Energy        (Qs and Ps for this section)       (back to the top)

As before, an object cannot do anything to itself to change the total amount of energy, but the energy can move between potential and kinetic.

The total energy may remain constant, but the potential and kinetic energies may not.  Think about you car coasting up a hill.  The higher it goes, the slower it goes.  (kinetic energy changing into potential energy)

Something from the outside of the object is needed to cause a change in the total energy.  That’s where work comes in:

                W = DTE  , but DTE could be the change in potential energy, or kinetic, or both.

EX. Consider shooting an arrow straight into the air.

• You pull back on the bowstring (work), increasing the potential energy of the string.
• When released, the string moves (potential changing to kinetic), pushing the arrow as it goes (work).
• The arrow moves at a fast speed as it comes off the bow (kinetic energy) which slows as the arrow goes up into the air (kinetic changing into potential).
• At the top of its flight the arrow is stopped (the total energy is all potential).
• As the arrow falls the arrow speeds up (potential changing into kinetic).
• when the arrow hits the ground its speed is approximately the same as when it left the bow.(the total energy is all kinetic).

In theory, the work done by you on the bowstring shows up as the maximum potential energy of the arrow.  In reality, some of the energy is absorbed into the potential and kinetic energy of the molecules in the string and bow, and the arrow does work on the air as it rises.  Each of these reduces the total energy of the arrow, and hence, its maximum height is not as high as it couold have been.

EX.  If you use an average force of 150 N to pull a bowstring back a distance of 0.30 m, how
         high will a 0.050 Kg arrow rise vertically?

               W = DTE
               W = D(PE + KE)
                                DKE = 0 since the arrow starts from rest and is at rest at the top of the rise
               W = DPE
             F•D = m g h
            150 N (0.30 m) = 0.050 Kg (9.8 m/s² )(h)
                h = 91.8 m
 

Conservation of Energy      (Qs and Ps for this section)       (back to the top)

As with momentum, energy is conserved in all interactions.  The general rule of the universe is that energy cannot be created or destroyed, just changed from one form to another.  In the early twentieth century this definition was expanded to include mass as well as energy since they are two forms of the same thing.  In this course we will only deal with energy conservation and assume the mass stays the same.

Elasticity is a term which refers to the amount of mechanical energy conserved in an interaction between objects.  If there is no loss of mechanical energy then the interaction is said to be perfectly elastic.  A nice thought, but impossible in real life.  Some energy is always lost to heat, light , sound, or deformation of the material.

Think of dropping a ball.  If the ball were to bounce back up to the same height from which it was dropped, it didn’t lose any energy.  This would be a completely elastic bounce.  On the other hand, if the ball stuck to the floor when you dropped it, it lost all of its energy; a completely inelastic bounce.

EX.  What is the degree of elasticity of a 0.080 Kg ball which is released from a height of
         1.5 m and bounces back to a height of 1.1 m?

                        original PE = m g h = 0.80 Kg (9.8 m/s² ) (1.5 m) = 11.76 J
                            final PE  = m g h = 0.80 Kg (9.8 m/s² ) (1.1 m) = 8.264 J

                            elasticity = ratio of final PE to initial PE
                                          = 8.264 J / 11.76 J
                                          =  0.733
                           The ball has an elasticity of 73.3 %
 
 

Rollercoasters are a great example of energy changes, or rather exchanges.  In general, some kind of device does work on the train to pull it up to the top of the first hill.  At this point the kinetic energy of the train is small due to its very low speed.  For the rest of the ride the train’s potential and kinetic energies are exchanged while the total energy remains the same, except for the energy removed by the negative work done by friction.

As the train decends from the top of the first hill its potential energy decreases with  a corresponding increase in kinetic energy which is why the train’s speed increases.  Usually the first drop of a rollercoaster contains the biggest transfer of energy from the potential to the kinetic so the train will go the fastest at the bottom of the first hill.

EX.    a)  The train for a rollercoaster sits in the loading house 5 m above the ground.
                If the train and its occupant’s combined mass is 1100 Kg, what is the potential energy
                of the train in the loading house?

                    PE = mgh = 1100 Kg (9.8 m/s² )(5 m) = 53900 J

          b)  After rolling to the bottom of the first incline the “pachink-pachink” machine pulls
               the train to the top of the first hill, 35 m high.  What is the change in potential energy
               of the train?

                DPE = mgh₂ - mgh₁ = mg(h₂ - h₁)
                                                  = 1100 Kg (9.8m/s² )(35m - 5 m)
                                                  = 323400 J

          c)  If it takes 20 sec to get the train to the top of the hill, what has to be the power of the
               “pachink-pachink” machine?

                   P = W / T = DPE / T = 323400 J / 20 s = 16170 w = 16.2 Kw

          d)   If the train is moving at 1.0 m/s at the top of the hill, what is its KE?

                   KE = 1/2 m V² = 1/2 (1100 Kg)(1 m/s)²  = 550 J

          e)  What is the total energy of the train at the top  of the hill?

                   TE = PE + KE = PE_{loading house} + DPE_{pachink machine} + KE_top

                         = 550 J + 53900 J + 323400 J = 377850 J
 

          f)  If the bottom of the first hill is at ground level, how much kinetic energy will the
               train have at that point?

                   The total energy of the train does not change while going down the first incline
                           KE = TE - PE = 377850 J - 0 = 377850 J

          g)  What is the speed of the train at the bottom of the first hill?

                   KE = 1/2 m V²         377850 J = 1/2 (1100 Kg)(V²)
                                                   26.2 m/s = V

          h)  If the train goes to the top of the next hill which is 15 m high, what is the train’s
               total energy at the top of this second hill?

                 TE = 377850 J   (the total energy doesn’t change except for a little lost to friction)

           i)  What is the potential energy of the train at the top of this second hill?

                  PE = mgh = 1100 Kg (9.8 m/s²)(15 m) = 161700 J

           j)  What is the KE of the train at the top of the second hill?

                  KE = TE - PE = 377850 J - 161700 J = 216150 J

           k)  How fast is the train moving at the top of the second hill?

                  KE = 1/2 m V²         216150 J = 1/2 (1100 Kg)(V²)
                                                  19.8 m/s = V
 

If you look at the equations for kinetic and potential energies you can equate the potential energy at the top of the hill to the kinetic energy at the bottom.

The mass drops out of the equation and the speed at the bottom of the hill would be:

Notice that the mass of the train is not involved.  An empty train would have the same speed at the bottom of the hill as a full one.  Interesting, eh?

In fact if you generalize the speed equation and let h be the change in height |h2 - h1|, then you could calculate the speed of the train at any point in the ride by only knowing the height of that point.  Try it for the top of the second hill.

                     V = sqrt(2(9.8 m/s²)|35 m - 15 m |) =  20 m/s

The same result as before.  Playing this potential/kinetic exchange can be a short cut to solving some problems.  You need little information to get an answer.
 

Elastic Collisions                                                (back to the top)

You have already worked problems involving collisions between two objects in your study of momentum.  Momentum is always conserved in any collision, but this fact does not fully explain why the outcome of a collision is always the same.  There are many combinations of final velocities which will satisy the conservation of momentum theorem.

When conservation of momentum is aided by the degree to which the kinetic energy is conserved unique solutions to collisions can be determined.  The elasticity of a collision is used to express the amount of mechanical kinetic energy conserved in an interaction.

If the objects in a collision stick together after the collision so they move with the same velocity the collision is defined as being perfectly inelastic.  There is still kinetic energy in the combination after the collision.  In fact the amount of the energy could range from near zero to almost as much as the kinetic energy they had before the collision.

EX.  Chocko the clown catches cannon balls in the circus.  If Chocko (80 Kg) catches a 7 Kg
         cannon ball moving at 30 m/s,

 a)  what is the speed of Chocko and the ball after the catch?
 b)  what is the percentage loss of kinetic energy in the catch?
 c)  where did the energy go?

                a)  P_T = P_T’                          m_cV_c  +  m_bV_b = m_cbV_cb’
                                            80 Kg (0)  +  7 Kg (30 m/s) =  87 Kg (V_cb’)
                                                                          2.4 m/s = V_cb’

                b)  KE before = 1/2(80 Kg)(0)² + 1/2(7 Kg)(30 m/s)² =  3150 J
                     KEafter   = 1/2(87 Kg)(2.4 m/s)² = 251 J
                          % lost = (3150 J - 251 J) / 3150 J = 0.92
                     92% of the energy was “lost”

                c) The “lost” kinetic energy would show up as increased energy in Chocko
                    (increased temperature)  and sound.
 

The other extreme of an elastic collision would be for it to be perfectly elastic.  In this case there is no loss of kinetic energy after the collision; kinetic energy is conserved.  As mentioned before no interaction is perfectly elastic, but some, like interactions between atoms or subatomic particles, come close.

EX.  A 0.6 Kg cue ball has a completely elastic collision with the 0.4 Kg 8-ball.  The cue ball is
        moving at 0.20 m/s and the 8-ball is moving in the opposite direction at 0.30 m/s before the
        collision.  What are the resulting velocities of each ball after the collision?

             The solution for this problem is algebraicly cluttered.  In practice it combines the equation
             for the conservation of momentum and kinetic energy, then solves for the velocities of the
             two balls.
 

                   •  For the conservation of momentum
                            0.6 Kg(.2 m/s)  +  0.4 Kg(-.3 m/s) = 0.6 Kg(V_c) + 0.4(V₈)
                                                                            0 = 0.6 Kg(V_c) + 0.4(V₈)

                   •  For the conservation of kinetic energy
             1/2(0.6 Kg)(.2 m/s)² + 1/2(0.4 Kg)(-.3 m/s)² = 1/2(0.6 Kg)(V_c)² + 1/2(0.4 Kg)(V₈)²
                                                                     0.03 J = 1/2(0.6 Kg)(V_c)² + 1/2(0.4 Kg)(V₈)²

                   •  Solve the equations by the substitution method
                             from the momentum equation:  Vc = -(0.67)V₈

                             substituting into the energy equation:
                                   0.03 J = 1/2(0.6 Kg)(V_c)² + 1/2(0.4 Kg)(V₈)²
                                   0.03 J = 1/2(0.6 Kg)(-(0.67)V₈)² + 1/2(0.4 Kg)(V₈)²

                   •  Simplifying this:
                                   0.03 J = 0.13 V₈²  + 0.2 V₈²
                                0.30 m/s = V₈

                   •  Since  V_c = -(0.67)V₈, then  V_c = -.02 m/s

       You can see that the solution is straight forward, but algebraically involved.
 

For collisions which are neither completely elastic or inelastic the solution is much the same as the previous one, but the total kinetic energy after the collision is a specific percentage of the initial kinetic energy.  This percentage would have to be known to solve the problem.  We will not attempt another solution of this type of problem at this time.  Once is enough.