Science notes, chapter 7
Energy and power
Textbook Science Form 3 (KSSM, Dual Language Programme), pages 208–229
What this chapter covers
- Define work and energy, and solve problems on work done in daily life (7.1.1)
- Relate power to work and solve problems on power in daily life (7.1.2)
- Explain gravitational potential energy and calculate it with E = mgh (7.2.1)
- Calculate elastic potential energy with E = ½Fx (7.2.2)
- Explain kinetic energy and calculate it with E = ½mv² (7.2.3)
- State the Principle of Conservation of Energy and explain it with oscillating systems (7.3.1)
- Solve problems on the transformation of kinetic and potential energy in a closed system (7.3.2)
7.1Work, energy and power
Work, W, is the product of the force, F, and the displacement, s, in the direction of the force.
Work (J) = Force (N) × Displacement (m), W = Fs
Work (J) = Force (N) × Displacement (m), W = Fs
The S.I. unit of work is the joule (J). 1 J of work is done when a force of 1 N moves an object through 1 m in the direction of the force, so 1 J = 1 N m. Larger units are the kilojoule (1 kJ = 1 000 J) and the megajoule (1 MJ = 1 000 000 J).
Displacement is the distance travelled in a specified direction. If the object does not move in the direction of the force, no work is done in that direction.
Worked examples
Lifting. A student weighing 400 N carries a load of 100 N up stairs of vertical height 3 m. The upward force needed is 400 + 100 = 500 N, so W = 500 × 3 = 1 500 J.Mass given. Lifting a box of mass 10 kg through 2 m (take 1 kg = 10 N): weight = 10 × 10 = 100 N, so W = 100 × 2 = 200 J.
Centimetres. Pulling a drawer 30 cm with a force of 2 N: 30 cm = 0.3 m, so W = 2 × 0.3 = 0.6 J.
Using the distance in cm. Change the displacement to metres before multiplying, otherwise the answer is not in joules. When the mass is given, first change it to weight (mass × 10).
Energy and power
Energy is the ability to do work. Its S.I. unit is the joule (J).
Power, P, is the rate of doing work.
Power (W) = Work done (J) ÷ Time taken (s), P = W ÷ t
Power, P, is the rate of doing work.
Power (W) = Work done (J) ÷ Time taken (s), P = W ÷ t
The S.I. unit of power is the watt (W): 1 W = 1 J s−1, meaning 1 J of work is done in 1 s. Two people who do the same work have different power if one does it in a shorter time. An old unit of power is the horsepower (hp).
Worked examples (textbook Table 7.2)
Monkey: weight 50 N, climbs 3 m in 20 s. W = 50 × 3 = 150 J; P = 150 ÷ 20 = 7.5 W.Box on a ramp: 30 N over 2 m in 5 s. W = 30 × 2 = 60 J; P = 60 ÷ 5 = 12 W.
Weight lifted: 150 N through 1 m in 0.5 s. W = 150 J; P = 150 ÷ 0.5 = 300 W.
Same work, less time = more power. To find power from a time in minutes, change the time to seconds first (1 minute = 60 s).
A crane lifts a load of 2 500 N through 4 m in 1.2 minutes. Find the work done, the energy used and the power.
Show answerHide answer
W = 2 500 × 4 = 10 000 J. Energy used = work done = 10 000 J. Time = 1.2 × 60 = 72 s, so P = 10 000 ÷ 72 = 138.9 W (about 139 W).
7.2Potential energy and kinetic energy
Gravitational potential energy
Gravitational potential energy is the work done to lift an object of mass m to a height h above the Earth's surface.
Ep = mgh (m in kg, g = 10 m s−2 or 10 N kg−1, h in m)
Ep = mgh (m in kg, g = 10 m s−2 or 10 N kg−1, h in m)
Weight = mass × gravitational acceleration. Lifting the object needs a force equal to its weight, mg, through a height h, so work done = mgh. With no other energy produced, all the work done becomes gravitational potential energy.
Photograph: Science Form 3 (DLP) textbook, Photograph 7.2, p. 216.
Worked example (textbook p. 217)
A lift carries a load of 1 500 kg to a height of 30 m in 0.5 minute.Work done = Ep = mgh = 1 500 × 10 × 30 = 450 000 J.
Time = 0.5 × 60 = 30 s, so P = 450 000 ÷ 30 = 15 000 W = 15 kW.
Elastic potential energy
Elastic potential energy is the work done to compress or stretch an elastic material (such as a spring) through a displacement x from its equilibrium position.
Ee = ½ Fx (F in N, x in m)
Ee = ½ Fx (F in N, x in m)
The force on a spring grows from 0 to F as it is stretched, so the average force is ½F. On a force–displacement graph the work done is the area under the graph, which is a triangle with area ½ × F × x.
Worked example (textbook p. 219)
Spring S has an original length of 20 cm and is compressed to 12 cm by a final force of 20 N.x = 20 − 12 = 8 cm = 0.08 m. Ee = ½ × 20 × 0.08 = 0.8 J.
The distance x is the change in length (original − new), not the new length, and it must be in metres.
Kinetic energy
Kinetic energy is the energy possessed by a moving object.
Ek = ½ mv2 (m in kg, v in m s−1)
Ek = ½ mv2 (m in kg, v in m s−1)
A stationary object has no kinetic energy. Because v is squared, doubling the speed makes the kinetic energy four times larger. This is why a heavy lorry at low speed and a small bullet at high speed can both have large kinetic energy.
| Quantity | To change units |
|---|---|
| Speed in km h−1 to m s−1 | Divide by 3.6 (× 1 000 m ÷ 3 600 s). 360 km h−1 = 100 m s−1 |
| Height or extension in cm | Divide by 100 to get metres |
| Mass in g | Divide by 1 000 to get kilograms |
Worked examples (textbook p. 220)
Train: mass 500 000 kg at 360 km h−1 = 100 m s−1. Ek = ½ × 500 000 × 1002 = 2 500 000 000 J.Find the speed: a 0.2 kg ball bearing has Ek = 3.6 J. 3.6 = ½ × 0.2 × v2, so v2 = 3.6 ÷ 0.1 = 36 and v = 6 m s−1.
Standard form: an electron of mass 9 × 10−31 kg at 4 × 106 m s−1: Ek = ½ × 9 × 10−31 × (4 × 106)2 = 7.2 × 10−18 J.
| Form | Formula | What the object has |
|---|---|---|
| Gravitational potential energy | mgh | A raised position above the Earth's surface |
| Elastic potential energy | ½Fx | A stretched or compressed elastic material |
| Kinetic energy | ½mv2 | Motion |
Liza lifts a chair weighing 40 N to a height of 50 cm. How much energy does the chair gain, and in what form?
Show answerHide answer
h = 0.5 m. Work done = 40 × 0.5 = 20 J, stored as gravitational potential energy (20 J).
7.3Principle of Conservation of Energy
The Principle of Conservation of Energy states that energy cannot be created or destroyed, but can only be converted from one form to another.
Oscillating systems such as a simple pendulum and a loaded spring keep transforming energy between potential energy (gravitational or elastic) and kinetic energy. A closed system has no external force such as friction, so no energy is changed to heat: the total of potential and kinetic energy stays constant.
Simple pendulum
| Position | Gravitational P.E. | K.E. | Why |
|---|---|---|---|
| X or Z (highest) | Maximum | Zero | Highest point, bob momentarily at rest |
| X to Y or Z to Y | Decreasing | Increasing | Bob falls and speeds up |
| Y (lowest) | Minimum | Maximum | Lowest height, greatest speed |
| Y to Z or Y to X | Increasing | Decreasing | Bob rises and slows down |
Loaded spring
| Position of the load | Elastic P.E. | K.E. |
|---|---|---|
| Most stretched or most compressed | Maximum | Zero (at rest) |
| Equilibrium position | Minimum | Maximum |
| Moving towards the equilibrium position | Decreasing | Increasing |
| Moving away from the equilibrium position | Increasing | Decreasing |
At the equilibrium position all the energy is kinetic; at the end points all of it is potential. In between, P.E. + K.E. = the same total.
Thinking energy is "used up" in a real pendulum. A real pendulum slows down because some energy changes to heat by friction and air resistance. The total energy is still conserved, but the heat is wasted energy, not useful energy.
Solving problems by conservation of energy
Toy pistol (textbook p. 226)
A spring is compressed by 250 mm (0.25 m) by a force of 5 N and fires a 50 g (0.05 kg) ball.Elastic P.E. = kinetic energy of the ball: ½ × 5 × 0.25 = ½ × 0.05 × v2.
0.625 = 0.025 v2, so v2 = 25 and v = 5 m s−1.
Assumption: no energy is lost to the surroundings.
Falling object
A 2 kg metal sphere is released from 2.5 m. Ep = mgh = 2 × 10 × 2.5 = 50 J. Just before it hits the ground all of it is kinetic: ½ × 2 × v2 = 50, so v2 = 50 and v = 7.1 m s−1 (√50 ≈ 7.07).A 40 g pendulum bob (0.04 kg) is released at X, 5 cm (0.05 m) above the lowest point Y. What is the gain in kinetic energy by the time it reaches Y?
Show answerHide answer
Loss of P.E. = mgh = 0.04 × 10 × 0.05 = 0.02 J. In a closed system this becomes kinetic energy, so the bob has 0.02 J of K.E. at Y (its maximum).