**Conservative and non-conservative forces **

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

When the only forces doing work are conservative forces (for example, gravitationalĀ and spring forces), energy changes forms – from kinetic to potential (or vice versa); yetĀ the total amount of mechanical energy (E_{K}+ E_{p}) is conserved. For example, as anĀ object falls in a gravitational ļ¬eld from a high elevation to a lower elevation, some ofĀ the objectās potential energy is changed into kinetic energy. However, the sum of theĀ kinetic and potential energies remain constant.

**Investigation: Non-conservative forces**

We can investigate the effect of non-conservative forces on an objectās total mechanicalĀ energy by rolling a ball along the ļ¬oor from point A to point B.

In the absence of friction and other non-conservative forces, the ball should slide alongĀ the ļ¬oor and its speed should be the same at positions A and B. Since there are noĀ non-conservative forces acting on the ball, its total mechanical energy at points A andĀ B are equal.

Now, letās investigate what happens when there is friction (an non-conservative force)Ā acting on the ball.

Roll the ball along a rough surface or a carpeted ļ¬oor. What happens to the speed ofĀ the ball at point A compared to point B?Ā

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

Letās compare the total mechanical energy of the ball at points A and B:

However, in this case, V_{A} ā V_{B} and therefore E_{Total,A} ā E_{Total,B}. Since

V_{A}>V_{B}

E_{Total,A}> E_{Total,B}

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 throughĀ applying a frictional force on the ball. In this case the carpet is doing negative work on

the ball.

When an non-conservative force (for example friction, air resistance, applied force)Ā does work on an object, the total mechanical energy (E_{K}+ E_{p}) Ā of that object changes.Ā If positive work is done, then the object will gain energy. If negative work is done,Ā then the object will lose energy.

When a net **force does work** on an object, then there is **always a change in the kineticĀ ****energy** of the object. This is because the object experiences an acceleration andĀ therefore a change in velocity.Ā This leads us to the work-energy theorem.

**DEFINITION**: Work-energy theorem

The work-energy theorem states that the work done on an object by the net force isĀ equal to the change in its kinetic energy:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā W_{Net}=ĪE_{k}=E_{k,f}-E_{k,i}

**Worked example 6: Work-energy theorem**

**QUESTION**

A 1 kg brick is dropped from a height of 10 m. Calculate the work that has been done on the brick between the moment it is released and the moment when it hits the ground. Assume that air resistance can be neglected.

**SOLUTION**

**Step 1: Determine what is given and what is required**

- Mass of the brick: m = 1 kg.
- Initial height of the brick: h
_{i}= 10 m.
- Final height of the brick: h
_{fi}= 0 m.
- We are required to determine the work done on the brick as it hits the ground.

**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 brickās

potential energy is converted to kinetic energy.

**Step 3: Determine the brickās potential energy at h**_{i}

Ep= m.g.h_{i}

Ā Ā Ā Ā Ā Ā Ā =(1) (9,8) (10)

= 98 J

**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.Ā When the brick hit the ground, it had 0 J of potential energy and 98 J of kinetic energy.

Therefore E_{k,i}= 0 J and E_{k,f}= 98 J.

From the work-energy theorem:

W_{net}Ā = ĪE_{k}

Ā Ā Ā Ā Ā = E_{k,f}–E_{k,i}

Ā Ā Ā Ā Ā = 98 – 0

Ā Ā Ā Ā Ā Ā = 98 J

Hence, 98 J of work was done on the brick.

**Worked example 7: Work-energy theorem 2**

**QUESTION**

The driver of a 1000 kg car travelling at a speed of 16,7 m.s^{-1Ā }applies the carās brakes when he sees a red light. The carās brakes provide a frictional force of 8000 N. Determine the stopping distance of the car.

**SOLUTION**

**Step 1: Determine what is given and what is required**

We are given:

- mass of the car: m = 1000 kg
- speed of the car: v = 16,7 m.s
^{-1}
- frictional force of brakes:Ā ā\( \tilde{F} \)ā= 8000 N

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

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

We apply the work-energy theorem. We know that all the carās kinetic energy is lostĀ to friction. Therefore, the change in the carās kinetic energy is equal to the work doneĀ by the frictional force of the carās brakes.Therefore, we ļ¬rst need to determine the carās kinetic energy at the moment of brakingĀ using: E_{k}=Ā½mv^{2}

This energy is equal to the work done by the brakes. We have the force applied by theĀ brakes, and we can use:Ā Ā Ā W = FĪx cos*Īø*Ā Ā to determine the stopping distance.

**Worked example 8: Block on an inclined plane [credit: OpenStax College]**

**QUESTION**

A block of 2 kg is pulled up along a smooth incline of length 10 m and height 5 m byĀ applying an non-conservative force. At the end of incline, the block is released fromĀ rest to slide down to the bottom. Find the

- work done by the non-conservative force,
- the kinetic energy of the block at the end of round trip, and
- the speed at the end of the round trip.

We have represented the non-conservative force on the force diagram with an arbitraryĀ vector\( \tilde{F} \)āĀ acts only during upward journey. Note that the block is simply released at the end of the upward journey. We need to ļ¬nd the work done by the non-conservativeĀ force only during the upward journey.

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā W_{F}=W_{F(up)}+W_{F(down)}=W_{F(up)}+0=W_{F(up)}

The kinetic energies in the beginning and at the end of the motion up the slope areĀ zero.

We can conclude that sum of the work done by all three forces is equal to zero duringĀ the upward motion. The change in kinetic energy is zero which implies that the netĀ work done is zero.

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā W_{NetĀ }= W_{F(up)}+W_{g(up)}+W_{N(up)}

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā 0Ā = W_{F(up)}+W_{g(up)}+W_{N(up)}

If we know the work done by the other two forces (normal force and gravity), then weĀ can calculate the work done by the non-conservative force, F, as required.

**Step 3: Work done by normal force during upward motion**

The block moves up the slope, the normal force is perpendicular to the slope and,

therefore, perpendicular to the direction of motion. Forces that are perpendicular toĀ the direction of motion do no work.Ā

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā W_{NetĀ }= W_{F(up)}+W_{g(up)}+W_{N(up)}

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā 0Ā = W_{F(up)}+W_{g(up)}+W_{N(up)}

Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā W_{F(up)}Ā = -W_{g(up)}

**IMPORTANT!**

Be careful not to be confused by which angle has been labelledĀ Ī± and which Īø.Ā Ī±Ā is not the angle between the force and the direction of motion but the incline of theĀ plane in this particular problem. It is important to understand which symbol representsĀ which physical quantity in the equations you have learnt.

Hence, the work done by the non-conservative force during the round trip isĀ

Ā Ā Ā Ā W_{F(up)}Ā = Ā W_{F(up)}= – W_{g(up)}

= – (-98)

= 98 J

**Step 5: Kinetic energy at the end of round trip**

The kinetic energy at the end of the upward motion was zero but it is not zero at theĀ end of the entire downward motion.

We can use the work-energy theorem to analyse the whole motion:

Ā Ā Ā Ā W_{(round trip)}Ā =Ā E_{k,f}– E_{k,i}

= E_{k,f}– 0

= E_{k,f}

**Exercise 5 ā 2: Energy**

1. Fill in the table with the missing information using the positions of the 1 kg ball in the diagram below combined with the work-energy theorem.

2. A falling ball hits the ground at 10 m.s^{-1}Ā 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 300 g is attached to the ceiling. It is pulled up to point A which is a height h = 30 cm from the equilibrium position.

Calculate the speed of the pendulum when it reaches point B (the equilibrium point). Assume that there are no non-conservative forces acting on the pendulum.