**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 force or an non-conservative 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 the only forces doing work are conservative forces (for example, gravitational and spring forces), energy changes forms – from kinetic to potential (or vice versa); yet the total amount of mechanical energy (E_{K}+ E_{p}) is conserved. 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.

**Investigation: Non-conservative forces**

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

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

Now, let’s investigate what happens when there is friction (an non-conservative 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 non-conservative frictional force, then the ball should be moving much slower at point B than at point A.

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

However, in this case, V_{A} ≠ V_{B} and therefore E_{Total,A} ≠ 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 through 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) does work on an object, the total mechanical energy (E_{K}+ E_{p}) 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.

When a net **force does work** on an object, then there is **always a change in the kinetic ****energy** of the object. This is because the object experiences an acceleration and therefore a change in velocity. 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 is equal to the change in its kinetic energy:

W_{Net}=ΔE_{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 brick’s

potential energy is converted to kinetic energy.

**Step 3: Determine the brick’s potential energy at h**_{i}

Ep= m.g.h_{i}

=(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. 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} = ΔE_{k}

= E_{k,f}–E_{k,i}

= 98 – 0

= 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^{-1 }applies the car’s brakes when he sees a red light. The car’s 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: \( \tilde{F} \)= 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 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: E_{k}=½mv^{2}

This energy is equal to the work done by the brakes. We have the force applied by the brakes, and we can use: W = FΔx cos*θ* 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 by applying an non-conservative force. At the end of incline, the block is released from 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 arbitrary vector\( \tilde{F} \) acts only during upward journey. Note that the block is simply released at the end of the upward journey. We need to ﬁnd the work done by the non-conservative force only during the upward journey.

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 are zero.

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

W_{Net }= W_{F(up)}+W_{g(up)}+W_{N(up)}

0 = 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 we 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 to the direction of motion do no work.

W_{Net }= W_{F(up)}+W_{g(up)}+W_{N(up)}

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

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

**IMPORTANT!**

Be careful not to be confused by which angle has been labelled α and which θ. α is not the angle between the force and the direction of motion but the incline of the plane in this particular problem. It is important to understand which symbol represents which physical quantity in the equations you have learnt.

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

W_{F(up)} = 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 the end of the entire downward motion.

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

W_{(round trip)} = E_{k,f}– E_{k,i}

= E_{k,f}– 0

= E_{k,f}

**Exercise 5 – 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} 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.