3.1Work, energy, power and deformation
WORK, Energy AND Power
5 min read 11 parts Lesson 2 of 17 1 quiz Virtual lab
Lesson summary
3.1 Work, Energy and Power — AS PHYSICS 9702 notes: Define work, energy and power, derive kinetic and gravitational potential energy and apply conservation of energy and efficiency.Work done = the energy transferred when a force moves its point of application: W = Fs cos θ, in jo…
- θ is the angle between force and displacement
- Work is a scalar
- Work done on a body increases its energy
- Work done by a gas expanding at constant pressure
- Energy cannot be created or destroyed
- Energy lost in one store equals the energy gained in others
Section 1 of 11Work done
Work done
Work done
Work done the energy transferred when a force moves its point of application: W = Fs cos θ, in joules.
θ is the angle between force and displacement
- θ is the angle between force and displacement only the component F cos θ along the displacement does work.
Work is a scalar
- Work is a scalar and can be negative.
Work done on a body increases its energy
- Work done on a body increases its energy work done by a body decreases its energy.
Work done by a gas expanding at constant pressure
- W = FΔx = pAΔx = pΔV.
Section 2 of 11Work done and the angle
| Angle θ | Work done | Example |
|---|---|---|
| 0° | Fs (maximum, positive) | Pulling a box along the direction of motion |
| 90° | 0 | Gravity on a horizontally moving object; centripetal force |
| 180° | −Fs | Friction opposing motion |
Section 3 of 11Conservation of energy
1Energy cannot be created or destroyed
- Energy cannot be created or destroyed it is transferred from one form to another, so the total energy of a closed system is constant.
2Energy lost in one store equals the energy gained in others
- Energy lost in one store equals the energy gained in others including work done against resistive forces, for example mgh = ½mv2 + Ffrictiond.
3Dissipated energy
Dissipated energy energy transferred to the surroundings in a non-useful form, mostly as thermal energy due to friction or air resistance.
4On a slope of angle θ
- On a slope of angle θ the height gained after travelling distance d along the slope is d sin θ.
Section 4 of 11Kinetic energy
Kinetic energy
Kinetic energy the energy a body has due to its motion: Ek = ½mv2.
What the symbols mean: Ek = kinetic energy (J) · m = mass (kg) · v = speed (m s−1)
Derivation
- Derivation a constant resultant force F accelerates mass m from rest over distance s, so W = Fs = mas.
Using v2 = 2as
- Using v2 = 2as as = v2 ÷ 2, giving W = ½mv2.
Also Ek = p2 ÷ 2m.
Section 5 of 11Gravitational potential energy
1GPE
GPE the energy a mass has because of its position in a gravitational field.
2Derivation
- Derivation lifting mass m at constant speed needs a force equal to its weight mg, so work over height Δh is mgΔh, stored as GPE.
3ΔEp
ΔEp mgΔh
What the symbols mean: Ep = gravitational potential energy gained (J) · m = mass (kg) · g = 9.81 N kg−1 · h = change in height (m)
- valid only where g is constant, close to the Earth's surface.
4Falling from rest with no air resistance
- Falling from rest with no air resistance mgh = ½mv2, so v = √(2gh).
Section 6 of 11Efficiency
1Efficiency
Efficiency useful energy output ÷ total energy input × 100%.
2
Equivalently useful power output ÷ total power input × 100%.
3It is always less than 100% in real devices
- It is always less than 100% in real devices some energy is dissipated, and wasted energy = input − useful output.
4
Give efficiency as a ratio or as a percentage, as asked.
Section 7 of 11Power
1Power
Power the rate of doing work or of energy transfer: P = W ÷ t = ΔE ÷ t, in watts.
What the symbols mean: P = power (W = J s−1) · W or E = work done / energy transferred (J) · t = time (s) · F = force (N) · v = velocity (m s−1)
2Derivation of P = Fv
- P = W ÷ t = Fs ÷ t and s ÷ t = v, so P = Fv for a constant force in the direction of velocity.
3A vehicle at constant (maximum) speed
- A vehicle at constant (maximum) speed driving force equals the total resistive force, so power output = resistive force × speed.
4Kilowatt-hour
1000 W × 3600 s = 3.6 × 106 J, a unit of energy, not power.
Section 8 of 11Measuring the power of a person running upstairs
1
Measure the mass of the person with scales and calculate their weight mg.
2
Measure the height of one step with a metre rule and count the steps to find the total vertical height h.
3
Time the person running up the stairs with a stopwatch.
4
Calculate the work done against gravity, mgh.
5
Calculate the output power P = mgh ÷ t.
6
Repeat and average the time; the value is a minimum because dissipated energy and kinetic energy are ignored.
Section 9 of 11Energy stores and equations
| Energy | Equation | Symbols |
|---|---|---|
| Kinetic | Ek = ½mv2 | m in kg, v in m s-1 |
| Gravitational potential (uniform field) | ΔEp = mgΔh | g = 9.81 N kg-1, h in m |
| Elastic (Hooke's law) | E = ½Fx = ½kx2 | k in N m-1, x in m |
| Work by a gas | W = pΔV | p in Pa, V in m3 |
Section 10 of 11Key terms
| Term | Meaning |
|---|---|
| Work done | The product of force and displacement moved in the direction of the force. |
| Joule (J) | The work done when a force of 1 N moves its point of application 1 m in the direction of the force. |
| Kinetic energy | The energy a body has due to its motion. |
| Gravitational potential energy | The energy a mass has due to its position in a gravitational field. |
| Power | The rate of doing work. |
| Watt (W) | A rate of energy transfer of one joule per second. |
| Efficiency | The ratio of useful output to total input. |
| Dissipated energy | Energy transferred to the surroundings in a non-useful form. |
Section 11 of 11Exam technique
1
Using the full force when only the component along the displacement does work.
2
Forgetting that work can be negative.
3
Quoting a kilowatt-hour as a unit of power — it is a unit of energy.
4
Using ΔEp = mgΔh for large heights where g is not constant.