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Mechanical Energy: Conservation of energy

Module by: Free High School Science Texts Project. E-mail the author

Mechanical Energy

Tip:

Mechanical energy is the sum of the gravitational potential energy and the kinetic energy.

Mechanical energy, EMEM, is simply the sum of gravitational potential energy (EPEP) and the kinetic energy (EKEK). Mechanical energy is defined as:

EM = EP + EK EM = EP + EK
(1)
EM = EP + EK EM = m g h + 1 2 m v 2 EM = EP + EK EM = m g h + 1 2 m v 2
(2)

Tip:

You may see mechanical energy written as UU. We will not use this notation in this book, but you should be aware that this notation is sometimes used.

Conservation of Mechanical Energy

The Law of Conservation of Energy states:

Energy cannot be created or destroyed, but is merely changed from one form into another.

Definition 1: Conservation of Energy

The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely changed from one form into another.

So far we have looked at two types of energy: gravitational potential energy and kinetic energy. The sum of the gravitational potential energy and kinetic energy is called the mechanical energy. In a closed system, one where there are no external forces acting, the mechanical energy will remain constant. In other words, it will not change (become more or less). This is called the Law of Conservation of Mechanical Energy and it states:

The total amount of mechanical energy in a closed system remains constant.

Definition 2: Conservation of Mechanical Energy

Law of Conservation of Mechanical Energy: The total amount of mechanical energy in a closed system remains constant.

This means that potential energy can become kinetic energy, or vice versa, but energy cannot 'disappear'. The mechanical energy of an object moving in the Earth's gravitational field (or accelerating as a result of gravity) is constant or conserved, unless external forces, like air resistance, acts on the object.

We can now use the conservation of mechanical energy to calculate the velocity of a body in freefall and show that the velocity is independent of mass.

Show by using the law of conservation of energy that the velocity of a body in free fall is independent of its mass.

Tip:

In problems involving the use of conservation of energy, the path taken by the object can be ignored. The only important quantities are the object's velocity (which gives its kinetic energy) and height above the reference point (which gives its gravitational potential energy).

Tip:

In the absence of friction, mechanical energy is conserved and
E M before = E M after E M before = E M after
(3)

In the presence of friction, mechanical energy is not conserved. The mechanical energy lost is equal to the work done against friction.

Δ EM = E M before - E M after = work done against friction Δ EM = E M before - E M after = work done against friction
(4)

In general, mechanical energy is conserved in the absence of external forces. Examples of external forces are: applied forces, frictional forces and air resistance.

In the presence of internal forces like the force due to gravity or the force in a spring, mechanical energy is conserved.

The following simulation covers the law of conservation of energy.
run demo

Figure 1
Figure 1 (energy-skate-park-screenshot.png)

Using the Law of Conservation of Energy

Mechanical energy is conserved (in the absence of friction). Therefore we can say that the sum of the EPEP and the EKEK anywhere during the motion must be equal to the sum of the EPEP and the EKEK anywhere else in the motion.

We can now apply this to the example of the suitcase on the cupboard. Consider the mechanical energy of the suitcase at the top and at the bottom. We can say:

Figure 2
Figure 2 (PG10C3_009.png)

E M top = E M bottom E P top + E K top = E P bottom + E K bottom m g h + 1 2 m v 2 = m g h + 1 2 m v 2 ( 1 ) ( 9 , 8 ) ( 2 ) + 0 = 0 + 1 2 ( 1 ) ( v 2 ) 19 , 6 J = 1 2 v 2 39 , 2 = v 2 v = 6 , 26 m · s - 1 E M top = E M bottom E P top + E K top = E P bottom + E K bottom m g h + 1 2 m v 2 = m g h + 1 2 m v 2 ( 1 ) ( 9 , 8 ) ( 2 ) + 0 = 0 + 1 2 ( 1 ) ( v 2 ) 19 , 6 J = 1 2 v 2 39 , 2 = v 2 v = 6 , 26 m · s - 1
(5)

The suitcase will strike the ground with a velocity of 6,26 m·s-16,26m·s-1.

From this we see that when an object is lifted, like the suitcase in our example, it gains potential energy. As it falls back to the ground, it will lose this potential energy, but gain kinetic energy. We know that energy cannot be created or destroyed, but only changed from one form into another. In our example, the potential energy that the suitcase loses is changed to kinetic energy.

The suitcase will have maximum potential energy at the top of the cupboard and maximum kinetic energy at the bottom of the cupboard. Halfway down it will have half kinetic energy and half potential energy. As it moves down, the potential energy will be converted (changed) into kinetic energy until all the potential energy is gone and only kinetic energy is left. The 19,6J19,6J of potential energy at the top will become 19,6J19,6J of kinetic energy at the bottom.

Exercise 1: Using the Law of Conservation of Mechanical Energy

During a flood a tree truck of mass 100kg100kg falls down a waterfall. The waterfall is 5m5m high. If air resistance is ignored, calculate

  1. the potential energy of the tree trunk at the top of the waterfall.
  2. the kinetic energy of the tree trunk at the bottom of the waterfall.
  3. the magnitude of the velocity of the tree trunk at the bottom of the waterfall.

Figure 3
Figure 3 (PG10C3_005.png)

Solution
  1. Step 1. Analyse the question to determine what information is provided :
    • The mass of the tree trunk m=100kgm=100kg
    • The height of the waterfall h=5mh=5m. These are all in SI units so we do not have to convert.
  2. Step 2. Analyse the question to determine what is being asked :
    • Potential energy at the top
    • Kinetic energy at the bottom
    • Velocity at the bottom
  3. Step 3. Calculate the potential energy. :
    E P = m g h E P = ( 100 ) ( 9 , 8 ) ( 5 ) E P = 4900 J E P = m g h E P = ( 100 ) ( 9 , 8 ) ( 5 ) E P = 4900 J
    (6)
  4. Step 4. Calculate the kinetic energy. :

    The kinetic energy of the tree trunk at the bottom of the waterfall is equal to the potential energy it had at the top of the waterfall. Therefore KE=4 900JKE=4 900J.

  5. Step 5. Calculate the velocity. :

    To calculate the velocity of the tree trunk we need to use the equation for kinetic energy.

    E K = 1 2 m v 2 4900 = 1 2 ( 100 ) ( v 2 ) 98 = v 2 v = 9 , 899 . . . v = 9 , 90 m · s - 1 downwards E K = 1 2 m v 2 4900 = 1 2 ( 100 ) ( v 2 ) 98 = v 2 v = 9 , 899 . . . v = 9 , 90 m · s - 1 downwards
    (7)

Exercise 2: Pendulum

A 2kg2kg metal ball is suspended from a rope. If it is released from point AA and swings down to the point BB (the bottom of its arc):

  1. Show that the velocity of the ball is independent of it mass.
  2. Calculate the velocity of the ball at point BB.

Figure 4
Figure 4 (PG10C3_006.png)

Solution
  1. Step 1. Analyse the question to determine what information is provided :
    • The mass of the metal ball is m=2kgm=2kg
    • The change in height going from point AA to point BB is h=0,5mh=0,5m
    • The ball is released from point AA so the velocity at point, vA=0m·s-1vA=0m·s-1.

    All quantities are in SI units.

  2. Step 2. Analyse the question to determine what is being asked :
    • Prove that the velocity is independent of mass.
    • Find the velocity of the metal ball at point BB.
  3. Step 3. Apply the Law of Conservation of Mechanical Energy to the situation :

    As there is no friction, mechanical energy is conserved. Therefore:

    E MA = E MA E PA + E KA = E PA + E KA m g h A + 1 2 m ( v A ) 2 = m g h B + 1 2 m ( v B ) 2 m g h A + 0 = 0 + 1 2 m ( v B ) 2 m g h A = 1 2 m ( v B ) 2 E MA = E MA E PA + E KA = E PA + E KA m g h A + 1 2 m ( v A ) 2 = m g h B + 1 2 m ( v B ) 2 m g h A + 0 = 0 + 1 2 m ( v B ) 2 m g h A = 1 2 m ( v B ) 2
    (8)

    As the mass of the ball mm appears on both sides of the equation, it can be eliminated so that the equation becomes:

    g h A = 1 2 ( v B ) 2 g h A = 1 2 ( v B ) 2
    (9)
    2 g h A = ( v B ) 2 2 g h A = ( v B ) 2
    (10)

    This proves that the velocity of the ball is independent of its mass. It does not matter what its mass is, it will always have the same velocity when it falls through this height.

  4. Step 4. Calculate the velocity of the ball :

    We can use the equation above, or do the calculation from 'first principles':

    ( v B ) 2 = 2 g h A ( v B ) 2 = ( 2 ) ( 9 . 8 ) ( 0 , 5 ) ( v B ) 2 = 9 , 8 v B = 9 , 8 m · s - 1 ( v B ) 2 = 2 g h A ( v B ) 2 = ( 2 ) ( 9 . 8 ) ( 0 , 5 ) ( v B ) 2 = 9 , 8 v B = 9 , 8 m · s - 1
    (11)

Potential Energy

  1. A tennis ball, of mass 120g120g, is dropped from a height of 5m5m. Ignore air friction.
    1. What is the potential energy of the ball when it has fallen 3m3m?
    2. What is the velocity of the ball when it hits the ground?
    Click here for the solution
  2. A bullet, mass 50g50g, is shot vertically up in the air with a muzzle velocity of 200m·s-1200m·s-1. Use the Principle of Conservation of Mechanical Energy to determine the height that the bullet will reach. Ignore air friction.
    Click here for the solution
  3. A skier, mass 50kg50kg, is at the top of a 6,4m6,4m ski slope.
    1. Determine the maximum velocity that she can reach when she skies to the bottom of the slope.
    2. Do you think that she will reach this velocity? Why/Why not?
    Click here for the solution
  4. A pendulum bob of mass 1,5kg1,5kg, swings from a height A to the bottom of its arc at B. The velocity of the bob at B is 4m·s-14m·s-1. Calculate the height A from which the bob was released. Ignore the effects of air friction.
    Click here for the solution
  5. Prove that the velocity of an object, in free fall, in a closed system, is independent of its mass.
    Click here for the solution

Summary

  • The potential energy of an object is the energy the object has due to his position above a reference point.
  • The kinetic energy of an object is the energy the object has due to its motion.
  • Mechanical energy of an object is the sum of the potential energy and kinetic energy of the object.
  • The unit for energy is the joule (J).
  • The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be changed from one form into another.
  • The Law of Conservation of Mechanical Energy states that the total mechanical energy of an isolated system remains constant.
  • The table below summarises the most important equations:
Table 1
Potential Energy E P = m g h E P = m g h
Kinetic Energy E K = 1 2 m v 2 E K = 1 2 m v 2
Mechanical Energy EM = E K + E P EM = E K + E P

End of Chapter Exercises: Gravity and Mechanical Energy

  1. Give one word/term for the following descriptions.
    1. The force with which the Earth attracts a body.
    2. The unit for energy.
    3. The movement of a body in the Earth's gravitational field when no other forces act on it.
    4. The sum of the potential and kinetic energy of a body.
    5. The amount of matter an object is made up of.
    Click here for the solution
  2. Consider the situation where an apple falls from a tree. Indicate whether the following statements regarding this situation are TRUE or FALSE. Write only 'true' or 'false'. If the statement is false, write down the correct statement.
    1. The potential energy of the apple is a maximum when the apple lands on the ground.
    2. The kinetic energy remains constant throughout the motion.
    3. To calculate the potential energy of the apple we need the mass of the apple and the height of the tree.
    4. The mechanical energy is a maximum only at the beginning of the motion.
    5. The apple falls at an acceleration of 9,8 m·s-29,8m·s-2.
    Click here for the solution
  3. A man fires a rock out of a slingshot directly upward. The rock has an initial velocity of 15 m·s-115m·s-1.
    1. What is the maximum height that the rock will reach?
    2. Draw graphs to show how the potential energy, kinetic energy and mechanical energy of the rock changes as it moves to its highest point.
    Click here for the solution
  4. A metal ball of mass 200g200g is tied to a light string to make a pendulum. The ball is pulled to the side to a height (A), 10cm10cm above the lowest point of the swing (B). Air friction and the mass of the string can be ignored. The ball is let go to swing freely.
    1. Calculate the potential energy of the ball at point A.
    2. Calculate the kinetic energy of the ball at point B.
    3. What is the maximum velocity that the ball will reach during its motion?
    Click here for the solution
  5. A truck of mass 1,2tons1,2tons is parked at the top of a hill, 150m150m high. The truck driver lets the truck run freely down the hill to the bottom.
    1. What is the maximum velocity that the truck can achieve at the bottom of the hill?
    2. Will the truck achieve this velocity? Why/why not?
    Click here for the solution
  6. A stone is dropped from a window, 6m6m above the ground. The mass of the stone is 25g25g. Use the Principle of Conservation of Energy to determine the speed with which the stone strikes the ground. Click here for the solution

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