Every person, animal and device transfers energy. Much of that energy is supplied by electricity, which must be generated from other energy stores. Some of these are renewable but most are non-renewable.

The amount of kinetic energy of a moving object can be calculated using the equation:

Kinetic energy = ½ × mass × (velocity)^{2}

E_{k} = ½ m v^{2}

This is when:

- kinetic energy (E
_{k}) is measured in joules (J) - mass (m) is measured in kilograms (kg)
- velocity (v) is measured in metres per second (m/s)

An apple of mass 100 g falls from a tree. It reaches a speed of 6 m/s before it lands on the ground. What is the gain of kinetic energy of the apple?

E_{k} = ½ m v^{2}

E_{k} = ½ × 0.1 × 6^{2}

E_{k} = ½ × 0.1 × 36

E_{k} = 1.8 J

- Question
How much kinetic energy does a 30 kg dog have when it runs at 4 m/s?

E

_{k}= ½ m v^{2}E

_{k}= ½ × 30 × 4^{2}E

_{k}= ½ × 30 × 16E

_{k}= 240 J

The amount of elastic potential energy stored in a stretched spring can be calculated using the equation:

Elastic potential energy = ½ × spring constant × (extension)^{2}

E_{e} = ½ k e^{2}

This is when:

- elastic potential energy (E
_{e}) is measured in joules (J) - spring constant (k) is measured in newtons per metre (N/m)
- extension (e) is measured in metres (m)

Robert stretches a spring with a spring constant of 3 N/m until it is extended by 50 cm.

What is the elastic potential energy stored by the spring?

E_{e} = ½ k e^{2}

E_{e} = ½ × 3 × (0.5)^{2}

E_{e} = ½ × 3 × 0.25

E_{e} = 0.75 J

- Question
How much elastic potential energy does a spring store when it is compressed by 0.2 m if it has a spring constant of 5 N/m?

E

_{e}= ½ k e^{2}E

_{e}= ½ × 5 ×(0.2)^{2}E

_{e}= ½ × 5 × 0.04E

_{e}= 0.1 J

The amount of gravitational potential energy stored by an object at height can be calculated using the equation:

Gravitational potential energy = mass × gravitational field strength × height

E_{p} = mgh

This is when:

- gravitational potential energy (E
_{p}) is measured in joules (J) - mass (m) is measured in kilograms (kg)
- gravitational field strength (g) is measured in newtons per kilogram (N/kg)
- height (h) is measured in metres (m)

A 5 kg ball is at rest at the top of a tower. The tower is 56 m high. Given that g = 10 N/kg, how much gravitational potential energy has the ball gained?

E_{p} = mgh

E_{p} = 5 × 10 × 56

E_{p} = 2,800 J

- Question
How much gravitational potential energy does a 500 g book gain when it is lifted up 1.5 m onto a shelf?

E

_{p}= mghE

_{p}= 0.5 × 10 × 1.5E

_{p}= 7.5 J

For any of these equations you may need to **change the subject of the formula**.

The amount of energy transformed in an electrical appliance can be calculated using the equation:

energy transformed = power × time

E = Pt

This is when:

- energy transformed in the appliance (E) is measured in joules (J)
- power (P) is the power of the appliance measured in watts (W)
- time (t) is the time the appliance is switched on for measured in seconds (s)

- Question
How much energy is transformed in a 2 kW kettle in two minutes?

2 kW = 2,000 W and two minutes = 120 seconds

E = Pt

E = 2,000 × 120

E = 240,000 J

Scientists can use the conservation of energy to estimate future values. For example, when the 5 kg ball is dropped from the top of the tower, how fast was it travelling when it hit the ground?

Assume the ball is streamlined so that air resistance is minimal. If this is the case, all the energy from the gravitational store goes into the kinetic store and it can be said that:

mgh = ½ mv^{2}

In the above example:

mgh = 2,800 J

½ mv^{2} = 2,800

Changing the subject:

v =

Since, m = 5

v =

v = 33.5 m/s

Most energy transfers result in energy being dissipated. An electric motor is being used to winch up a lift. The energy transformed in the motor should increase the gravitational potential energy store but because the motor becomes hot during use, the thermal energy store of the surroundings is also increased as energy is dissipated.

The lift has a mass of 400 kg and is lifted up a vertical distance of 25 m. The power of the electric motor is 5,000 W and it takes the motor 25 s to move the lift. How much energy is dissipated to the surroundings?

First find the increase of the gravitational potential energy store:

E_{p} = mgh

Ep = 400 × 10 × 25

E_{p} = 100,000 J

Next find the energy transformed in the electric motor:

E = Pt

E = 5,000 × 25

E = 125,000 J

The energy transformed in the motor is more than the gravitational potential energy gained by the lift. The difference has been dissipated as heat to the surroundings. The difference is:

125,000 - 100,000

= 25,000 J

So, 25,000 J has been dissipated to the surroundings.