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Energy

External and Internal Forces

In Grade 10, you saw that mechanical energy was conserved in the absence of external forces. It is important to know whether a force is an internal force or an external force in the system, because this is related to whether the force can change an object's total mechanical energy when it does work on an object.

When an external force (for example friction, air resistance, applied force) does work on an object, the total mechanical energy (KE + PE) of that object changes. If positive work is done, then the object will gain energy. If negative work is done, then the object will lose energy. The gain or loss in energy can be in the form of potential energy, kinetic energy, or both. However, the work which is done is equal to the change in mechanical energy of the object.

Investigations : External Forces

We can investigate the effect of external forces on an object's total mechanical energy by rolling a ball along the floor from point A to point B.

Figure 1
Figure 1 (PG12C3_012.png)

Find a nice smooth surface (e.g. a highly polished floor), mark off two positions, A and B, and roll the ball between them.

The total mechanical energy of the ball, at each point, is the sum of its kinetic energy (KE) and gravitational potential energy (PE):

E total , A = KE A + PE A = 1 2 m v A 2 + m g h A = 1 2 m v A 2 + m g ( 0 ) = 1 2 m v A 2 E total , A = KE A + PE A = 1 2 m v A 2 + m g h A = 1 2 m v A 2 + m g ( 0 ) = 1 2 m v A 2
(1)
E total , B = KE B + PE B = 1 2 m v B 2 + m g h B = 1 2 m v B 2 + m g ( 0 ) = 1 2 m v B 2 E total , B = KE B + PE B = 1 2 m v B 2 + m g h B = 1 2 m v B 2 + m g ( 0 ) = 1 2 m v B 2
(2)

In the absence of friction and other external forces, the ball should slide along the floor and its speed should be the same at positions A and B. Since there are no external forces acting on the ball, its total mechanical energy at points A and B are equal.

v A = v B 1 2 m v A 2 = 1 2 m v B 2 E total , A = E total , B v A = v B 1 2 m v A 2 = 1 2 m v B 2 E total , A = E total , B
(3)

Now, let's investigate what happens when there is friction (an external force) acting on the ball.

Roll the ball along a rough surface or a carpeted floor. What happens to the speed of the ball at point A compared to point B?

If the surface you are rolling the ball along is very rough and provides a large external frictional force, then the ball should be moving much slower at point B than at point A.

Let's now compare the total mechanical energy of the ball at points A and B:

E total , A = KE A + PE A = 1 2 m v A 2 + m g h A = 1 2 m v A 2 + m g ( 0 ) = 1 2 m v A 2 E total , A = KE A + PE A = 1 2 m v A 2 + m g h A = 1 2 m v A 2 + m g ( 0 ) = 1 2 m v A 2
(4)
E total , B = KE B + PE B = 1 2 m v B 2 + m g h B = 1 2 m v B 2 + m g ( 0 ) = 1 2 m v B 2 E total , B = KE B + PE B = 1 2 m v B 2 + m g h B = 1 2 m v B 2 + m g ( 0 ) = 1 2 m v B 2
(5)

However, in this case, vAvBvAvB and therefore E total ,AE total ,BE total ,AE total ,B. Since

v A > v B E total , A > E total , B v A > v B E total , A > E total , B
(6)

Therefore, the ball has lost mechanical energy as it moves across the carpet. However, although the ball has lost mechanical energy, energy in the larger system has still been conserved. In this case, the missing energy is the work done by the carpet through applying a frictional force on the ball. In this case the carpet is doing negative work on the ball.

When an internal force does work on an object by an (for example, gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant but the object's energy can change form. For example, as an object falls in a gravitational field from a high elevation to a lower elevation, some of the object's potential energy is changed into kinetic energy. However, the sum of the kinetic and potential energies remain constant. When the only forces doing work are internal forces, energy changes forms - from kinetic to potential (or vice versa); yet the total amount of mechanical energy is conserved.

Capacity to do Work

Energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system. An object with larger potential energy has a greater capacity to do work.

Exercise 1: Work Done on a System

Show that a hammer of mass 2 kg does more work when dropped from a height of 10 m than when dropped from a height of 5 m. Confirm that the hammer has a greater potential energy at 10 m than at 5 m.

Solution
  1. Step 1. Determine what is given and what is required :

    We are given:

    • the mass of the hammer, m=m=2 kg
    • height 1, h1h1=10 m
    • height 2, h2h2=5 m

    We are required to show that the hammer does more work when dropped from h1h1 than from h2h2. We are also required to confirm that the hammer has a greater potential energy at 10 m than at 5 m.

  2. Step 2. Determine how to approach the problem :
    1. Calculate the work done by the hammer, W1W1, when dropped from h1h1 using:
      W1=Fg·h1.W1=Fg·h1.
      (7)
    2. Calculate the work done by the hammer, W2W2, when dropped from h2h2 using:
      W2=Fg·h2.W2=Fg·h2.
      (8)
    3. Compare W1W1 and W2W2
    4. Calculate potential energy at h1h1 and h2h2 and compare using:
      PE=m·g·h.PE=m·g·h.
      (9)
  3. Step 3. Calculate W1W1 :
    W 1 = F g · h 1 = m · g · h 1 = ( 2 kg ) ( 9 . 8 m · s - 2 ) ( 10 m ) = 196 J W 1 = F g · h 1 = m · g · h 1 = ( 2 kg ) ( 9 . 8 m · s - 2 ) ( 10 m ) = 196 J
    (10)
  4. Step 4. Calculate W2W2 :
    W 2 = F g · h 2 = m · g · h 2 = ( 2 kg ) ( 9 . 8 m · s - 2 ) ( 5 m ) = 98 J W 2 = F g · h 2 = m · g · h 2 = ( 2 kg ) ( 9 . 8 m · s - 2 ) ( 5 m ) = 98 J
    (11)
  5. Step 5. Compare W1W1 and W2W2 :

    We have W1W1=196 J and W2W2=98 J. W1>W2W1>W2 as required.

  6. Step 6. Calculate potential energy :

    From Equation 9, we see that:

    P E = m · g · h = F g · h = W P E = m · g · h = F g · h = W
    (12)

    This means that the potential energy is equal to the work done. Therefore, PE1>PE2PE1>PE2, because W1>W2W1>W2.

This leads us to the work-energy theorem.

Definition 1: Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy:

W = Δ K E = K E f - K E i W = Δ K E = K E f - K E i
(13)

The work-energy theorem is another example of the conservation of energy which you saw in Grade 10.

Exercise 2: Work-Energy Theorem

A brick of mass 1 kg is dropped from a height of 10 m. Calculate the work done on the brick at the point it hits the ground assuming that there is no air resistance?

Solution
  1. Step 1. Determine what is given and what is required :

    We are given:

    • mass of the brick: mm=1 kg
    • initial height of the brick: hihi=10 m
    • final height of the brick: hfhf=0 m

    We are required to determine the work done on the brick as it hits the ground.

  2. Step 2. Determine how to approach the problem :

    The brick is falling freely, so energy is conserved. We know that the work done is equal to the difference in kinetic energy. The brick has no kinetic energy at the moment it is dropped, because it is stationary. When the brick hits the ground, all the brick's potential energy is converted to kinetic energy.

  3. Step 3. Determine the brick's potential energy at hihi :
    P E = m · g · h = ( 1 kg ) ( 9 , 8 m · s - 2 ) ( 10 m ) = 98 J P E = m · g · h = ( 1 kg ) ( 9 , 8 m · s - 2 ) ( 10 m ) = 98 J
    (14)
  4. Step 4. Determine the work done on the brick :

    The brick had 98 J of potential energy when it was released and 0 J of kinetic energy. When the brick hit the ground, it had 0 J of potential energy and 98 J of kinetic energy. Therefore KEiKEi=0 J and KEfKEf=98 J.

    From the work-energy theorem:

    W = Δ K E = K E f - K E i = 98 J - 0 J = 98 J W = Δ K E = K E f - K E i = 98 J - 0 J = 98 J
    (15)
  5. Step 5. Write the final answer :

    98 J of work was done on the brick.

Exercise 3: Work-Energy Theorem 2

The driver of a 1 000 kg car traveling at a speed of 16,7 m·s-1m·s-1 applies the car's brakes when he sees a red robot. The car's brakes provide a frictional force of 8000 N. Determine the stopping distance of the car.

Solution
  1. Step 1. Determine what is given and what is required :

    We are given:

    • mass of the car: mm=1 000 kg
    • speed of the car: vv=16,7 m·s-1m·s-1
    • frictional force of brakes: FF=8 000 N

    We are required to determine the stopping distance of the car.

  2. Step 2. Determine how to approach the problem :

    We apply the work-energy theorem. We know that all the car's kinetic energy is lost to friction. Therefore, the change in the car's kinetic energy is equal to the work done by the frictional force of the car's brakes.

    Therefore, we first need to determine the car's kinetic energy at the moment of braking using:

    K E = 1 2 m v 2 K E = 1 2 m v 2
    (16)

    This energy is equal to the work done by the brakes. We have the force applied by the brakes, and we can use:

    W = F · d W = F · d
    (17)

    to determine the stopping distance.

  3. Step 3. Determine the kinetic energy of the car :
    K E = 1 2 m v 2 = 1 2 ( 1 000 k g ) ( 16 , 7 m s ) 2 = 139 445 J K E = 1 2 m v 2 = 1 2 ( 1 000 k g ) ( 16 , 7 m s ) 2 = 139 445 J
    (18)
  4. Step 4. Determine the work done :

    Assume the stopping distance is d0d0. Then the work done is:

    W = F · d = ( - 8 000 N ) ( d 0 ) W = F · d = ( - 8 000 N ) ( d 0 )
    (19)

    The force has a negative sign because it acts in a direction opposite to the direction of motion.

  5. Step 5. Apply the work-enemy theorem :

    The change in kinetic energy is equal to the work done.

    Δ K E = W K E f - K E i = ( - 8 000 N ) ( d 0 ) 0 J - 139 445 J = ( - 8 000 N ) ( d 0 ) d 0 = 139 445 J 8 000 N = 17 , 4 m Δ K E = W K E f - K E i = ( - 8 000 N ) ( d 0 ) 0 J - 139 445 J = ( - 8 000 N ) ( d 0 ) d 0 = 139 445 J 8 000 N = 17 , 4 m
    (20)
  6. Step 6. Write the final answer :

    The car stops in 17,4 m.

Tip:

A force only does work on an object for the time that it is in contact with the object. For example, a person pushing a trolley does work on the trolley, but the road does no work on the tyres of a car if they turn without slipping (the force is not applied over any distance because a different piece of tyre touches the road every instant.

Tip:

Energy is conserved!

Tip: Energy Conservation::

In the absence of friction, the work done on an object by a system is equal to the energy gained by the object.

Work Done = Energy Transferred Work Done = Energy Transferred
(21)

In the presence of friction, only some of the energy lost by the system is transferred to useful energy. The rest is lost to friction.

Total work done = Useful work done + Work done against friction Total work done = Useful work done + Work done against friction
(22)

In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational field of the earth is what does the work in this case.

Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy.

Other forms of Potential Energy

  1. elastic potential energy - potential energy is stored in a compressed or extended spring or rubber band. This potential energy is calculated by:
    12kx212kx2
    (23)
    where kk is a constant that is a measure of the stiffness of the spring or rubber band and xx is the extension of the spring or rubber band.
  2. Chemical potential energy is related to the making and breaking of chemical bonds. For example, a battery converts chemical energy into electrical energy.
  3. The electrical potential energy of an electrically charged object is defined as the work that must be done to move it from an infinite distance away to its present location, in the absence of any non-electrical forces on the object. This energy is non-zero if there is another electrically charged object nearby otherwise it is given by:
    kq1q2dkq1q2d
    (24)
    where kk is Coulomb's constant. For example, an electric motor lifting an elevator converts electrical energy into gravitational potential energy.
  4. Nuclear energy is the energy released when the nucleus of an atom is split or fused. A nuclear reactor converts nuclear energy into heat.

Some of these forms of energy will be studied in later chapters.

Investigation : Energy Resources

Energy can be taken from almost anywhere. Power plants use many different types of energy sources, including oil, coal, nuclear, biomass (organic gases), wind, solar, geothermal (the heat from the earth's rocks is very hot underground and is used to turn water to steam), tidal and hydroelectric (waterfalls). Most power stations work by using steam to turn turbines which then drive generators and create an electric current.

Most of these sources are dependant upon the sun's energy, because without it we would not have weather for wind and tides. The sun is also responsible for growing plants which decompose into fossil fuels like oil and coal. All these sources can be put under 2 headings, renewable and non-renewable. Renewable sources are sources which will not run out, like solar energy and wind power. Non-renewable sources are ones which will run out eventually, like oil and coal.

It is important that we learn to appreciate conservation in situations like this. The planet has a number of linked systems and if we don't appreciate the long-term consequences of our actions we run the risk of doing damage now that we will only suffer from in many years time.

Investigate two types of renewable and two types of non-renewable energy resources, listing advantages and disadvantages of each type. Write up the results as a short report.

Figure 2

Energy

  1. Fill in the table with the missing information using the positions of the ball in the diagram below combined with the work-energy theorem.
    Figure 3
    Figure 3 (PG12C3_014.png)
    Table 1
    positionKEKEPEPEvv
    A 50 J 
    B 30 J 
    C   
    D 10 J 
    E   
    F   
    G   
  2. A falling ball hits the ground at 10 m··s-1-1 in a vacuum. Would the speed of the ball be increased or decreased if air resistance were taken into account. Discuss using the work-energy theorem.
  3. A pendulum with mass 300g is attached to the ceiling. It is pulled up to point A which is a height h = 30 cm from the equilibrium position.
    Figure 4
    Figure 4 (PG12C3_015.png)
    Calculate the speed of the pendulum when it reaches point B (the equilibrium point). Assume that there are no external forces acting on the pendulum.

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