WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Energy Concepts
The work–energy relationship
- 3 things to know
- AS Unit 1
Overview
Recall questions on The work–energy relationship.
What you need to know
Answer each one before you reveal it.
Write down the relationship between the work done by the resultant force on a body and the change in the body's kinetic energy.
What does it mean for the work done on a body to be negative?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Write down the relationship between the work done by the resultant force on a body and the change in the body's kinetic energy.
Model answer
Fx = ½mv² − ½mu², so the work done by the resultant force is equal to the gain in kinetic energy.
- 2
Define each symbol in the relationship Fx = ½mv² − ½mu², giving its unit.
Model answer
F is the resultant force in newtons, x the distance moved in metres, m the mass in kilograms, u the initial speed and v the final speed, both in m s⁻¹.
- 3
Show that the work–energy relationship follows from F = ma together with v² = u² + 2ax.
Model answer
Rearranging the equation of motion gives a = (v² − u²)/2x, and substituting that into F = ma gives F = m(v² − u²)/2x, so Fx = ½mv² − ½mu².
- 4
What does it mean for the work done on a body to be negative?
Model answer
The force has a component opposing the motion, so energy is taken from the body and its kinetic energy falls.
- 5
A 1200 kg car moving at 20 m s⁻¹ comes to a halt over a braking distance of 40 m. Use energy to find the mean braking force.
Model answer
The kinetic energy to be removed is ½ × 1200 × 20² = 2.4 × 10⁵ J, and dividing that by 40 m gives a mean braking force of 6.0 × 10³ N.
- 6
Why does doubling the speed of a car quadruple its braking distance for the same braking force?
Model answer
The kinetic energy to be removed is proportional to the square of the speed, and the distance is that energy divided by a fixed braking force.
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