Work, Energy and Power – SS2 Further Mathematics Lesson Note (Second Term)

Work is done when the point of application of a force is displaced. It is the product of the force and the distance moved along the line of the force. Power is the rate of doing work and is measured in watts. The kinetic energy of a particle of mass m moving with velocity v is ½mv², and its potential energy at height h is mgh.

Behavioural objectives

By the end of the lesson, students should be able to:

• define work done by a constant force and calculate it;
• define power and state its unit;
• define kinetic energy and potential energy and write their expressions;
• show that the change in kinetic energy is ½mv² – ½mu²;
• state the law of conservation of energy and use it to find the velocity of a freely falling particle on reaching the ground.

What is work?

If the point of application of a force is displaced, the force is said to do work. The work done by a constant force is the product of the force and the distance moved by its point of application along the line of application of the force.

W = |F| × d

Here |F| is the magnitude of the force and d is the magnitude of the displacement.

What is power?

Power is the rate at which work is being done. For example, if work of 90 J is done in 15 seconds, the power is 6 J/s. The unit of power is the watt (W).

Worked example: power of a car

On the level, a car develops a power of 60 kW. If the resistance to motion is 900 N, what is the maximum speed of the car?

Working at the same power and with the same resistance operating, what would be the maximum speed possible up an inclined plane whose slope is sin^-1(1/50), if the mass of the car is 800 kg?

What is the acceleration at the time when the car is moving up the inclined plane at 40 m s^-1? (Take g = 10 m s^-2)

Solution

Let P be the power developed by the car.

Let W be the work done.

Let F be the tractive force of the car.

P = dW/dt = d(F × s)/dt

What is energy?

What is kinetic energy?

The work done in bringing a particle of mass m from rest to a velocity v is called the kinetic energy of the particle.

If we denote by E_k the kinetic energy of a particle of mass m reaching a velocity v from rest, then

E_k = ½mv²

Now let W be the work done in bringing the particle of mass m to a velocity v from an initial velocity u, and let s be the distance travelled in the process. Then

W = F × s = m × a × s, where a is the acceleration.

But

s = (u + v)t/2

a = (v – u)/t

Therefore W = m × a × s

= m × (v – u)/t × (v + u)t/2

= ½mv² – ½mu²

If we denote the last expression by ΔE_k, then

ΔE_k = ½mv² – ½mu²

ΔE_k is the change in kinetic energy in bringing a particle of mass m from an initial velocity u to a final velocity v.

What is potential energy?

The potential energy of a particle of mass m is the energy the particle has due to its position relative to a reference level. It is the work done in bringing a particle from a reference level to a height h.

If we denote the potential energy by E_p, then

E_p = F × s

Since F = mg and s = h,

E_p = mgh

What is the law of conservation of energy?

The law is a generalisation from experience of nature. It states that energy can be neither created nor destroyed. It can only be changed from one form to another.

When a particle falls from a height, it loses potential energy. This loss in potential energy is compensated for by the gain in kinetic energy.

Worked example: velocity of a falling particle

A particle starting from rest falls freely from a height H above the ground. If g is the acceleration due to gravity, show from energy consideration that the velocity v with which the particle strikes the ground is given by the expression

v = √(2gH)

Solution

Let the particle have a mass m kg.

Loss of potential energy = mgH.

Gain in kinetic energy = ½mv².

The loss in potential energy equals the gain in kinetic energy, so mgH = ½mv². Hence v² = 2gH and v = √(2gH).

Evaluation

1. Define the work done by a constant force.
2. Define power and state its unit.
3. Write the expressions for the kinetic energy and the potential energy of a particle of mass m.
4. State the law of conservation of energy.

Assignment

1. A body of mass 20 kg moves a distance of 8 m in the direction of the line of action of a force F = 5 N on a smooth table. Find the work done.
2. A particle of mass 3 kg is projected vertically upward with an initial velocity of 5 m/s from the ground. Calculate the potential energy at the greatest height.
3. A sphere of mass 12 kg and another sphere of mass 8 kg move toward each other with velocities 5 m/s and 3 m/s respectively. Find the speed of the spheres after collision.
4. Calculate the loss in kinetic energy caused by the collision of the two bodies in question 3.