Work and Energy Conservation

 

Mechanical energy describes the ability of an object to do work. The work done on an object arises from a force applied over a distance (W=Fdd) which either accelerates the object thus changing its motional energy (kinetic energy), or stores energy by changing its position (potential energy). For instance, when a moving car is brought to rest, the work done by the frictional force on the tires is equal to the kinetic energy of the car, KE=1/2 mv2. In addition, forces which are provided by the car's engine can do work in climbing up a hill which is stored as gravitational potential energy, PE=mgh. The mechanical energy of an object is equal to the sum of the potential plus kinetic energies, i.e. E = PE + KE, and is a direct measure of the total energy available to an object as its speed and position changes from one point to another.

In special cases where energy is not lost to the environment due to nonconservative forces such as friction, the mechanical energy of a system of masses remains constant as the object or objects move in space. In other words, the energy of the system stays the same for all times in the life of the objects. Since there is no energy created or destroyed as the objects travel along on their paths, the forces which do work exchange energy between potential and kinetic according to the energy initially provided to the system. This is the basis for the law of  conservation of energy which can be written as E = PE + KE = constant. As an example, consider an apple of mass m which is initially a height h above the ground. When the apple is hanging from the tree, it stores an amount of gravitational potential energy as it is held motionless so that the initial mechanical energy is Etop = PEtop = mgh. If the apple is cut from the tree, forces due to gravity do work which cause it to fall and speed up at the same time. Just before it hits the ground, the potential energy is zero such that the mechanical energy arises solely due to the kinetic energy, Ebot = KEbot = 1/2 mv2. Using the conservation of energy for this frictionless problem, we can equate the mechanical energy at the top and bottom of the tree, or more specifically                                       E top =E top

PE top = KE bot

mgh = 1/2 mv2.

It is clear from the equations above that the mechanical energy allows you to relate the position of the apple to its velocity as it falls from the tree. As potential energy is lost from the decrease in height above the ground, kinetic energy is gained while the object speeds up.

 

Of course if we include all types of system energy including the internal heating of the object, and light or sound given off as it moves, we can apply the conservation of energy to cases involving friction or even collisional events. In such situations, we must be able to compute the work done by the nonconservative forces which tend to reduce the initial mechanical energy provided to the system. However, if the total energy of the system and environment are taken into consideration, we again see that energy is neither created nor destroyed. The details of problems involving frictional work will be left for a more advanced course in the subject.

 

Another case to consider involves an object traveling on a level ground which is brought to rest by an external force. Since the force is applied over a finite distance, then work is done on the mass which decelerates it to a stop. The total work required to bring the system to a halt is simply given by the initial kinetic energy; there is no change in potential energy if it travels at the same height on the ground so that the kinetic energy is removed from the object by the external forces. Alternatively, if a force is used to speed up a mass the work is equal to the kinetic energy gained by the object,

W = KE = 1/2 mv2.

 

In order to reinforce the concepts noted above, consider the following examples which explore the motion of everyday objects as they move in the presence of forces which change either the position or velocity of the masses.

 

EX. #1: A helicopter drops a 20 kg package from rest at a height of 120 m from the ground.

a) How much initial potential energy is stored by the package?

b) What is the kinetic energy of the package just before it hits the ground?

c) What is the speed of the package just before it hits the ground?

SOLUTION:

a)      The distance of the package from the ground is given as 120m. This provides a reference potential energy at the ground of zero since h=0 when the object hits the ground below the helicopter. If the zero of gravitational potential energy is defined this way, then the potential energy of the package can be written as

PEgrav=mgh = (20 kg)(10 m/s2)(120 m) = 24000 J

b)      By ignoring any energy lost due to friction, we can use the conservation of energy to solve for the energy after the package hits the ground. Since the package is held motionless above the ground, we see that the initial mechanical energy is simply the potential energy it has while in the helicopter (no KE):

Etop = PEtop + KEtop = 24000 J

As the object falls and hits the ground at the bottom, it loses all of the potential energy (reaches the point where h=0) which is transformed into kinetic:

Ebot = PEbot + KEbot = KEbot

Now applying the concept of energy conservation, we equate the energy at the top to the energy at the bottom to get the final answer

Ebot = Etop

KEbot = 24000 J

Notice that this simply says that all of the stored potential energy the package had while in the helicopter is transformed into kinetic just before it hits the ground.

c)      Since we know the definition of the kinetic energy KE=1/2 mv2, we just need to solve for the speed of the package by setting the answer above equal to the KE formula:

KE=1/2 mv2 =1/2 (20 kg) v2 = 24000 J

v2= 2400 m2/s2                        v = 48.99 m/s

 

EX. #2: A 5 kg cannonball is fired from the ground straight up into the air such that it rises to a height of 50 m.

a) What is the potential energy of the cannonball when it gets to its maximum height?

b) What is the speed of the cannonball just before it falls back to the ground? Note that if we ignore air friction, this is the same as the initial speed of the cannonball when it was fired from the cannon.

SOLUTION:   

a)      Again we define the zero of potential energy to be the lowest point in the problem, where the cannonball is initially fired. Then we can use the height of the cannonball above the ground to determine the potential energy from the formula:

PE = mgh = (5 kg)(10 m/s2)(50 m) = 2500 J

As in the problem above, when the cannonball hits the ground, all of the potential energy it had at the maximum height is transformed into kinetic energy. Using the same idea of energy conservation, we relate the potential energy lost by the cannonball to the kinetic gained: PE top = KE bot.

PE lost = KE gained= 1/2 mv2

(5 kg) v2 = 2500 J

v2= 1000 m2/s2                                    v= 31.62 m/s

 

EX. #3: A 2.5 kg bowling ball rolls into a wall traveling at 6 m/s. It crashes into the wall and is brought to rest by the forces acting from the wall.

a) How much work was done by the wall in bringing the object to rest?

b) If the ball comes to rest in 0.4 seconds, then calculate the power used in stopping the ball.

SOLUTION:   

a)      Since the bowling ball stops after hitting the wall, all of the kinetic energy it had initially is removed by the work due to forces by the wall. In this case the work done is equal to the change in kinetic energy of the ball, W=KE:

W= KE = 1/2mv2

W = (2.5 kg)(6 m/s)2 = 45 J

b) The power is defined as the rate of change in work or energy P = (Work Energy)/time. The energy used to stop the ball was found above from the change in kinetic energy, or work done by the wall in bringing the bowling ball to rest. Using the time to stop the ball in addition to the power formula, the power can be calculated as

P = (Work Energy)/time = 45 J/0.4 s = 112.5 W