Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
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
Work done when the force is not along the motion
- 3 things to know
- Component 1
Overview
Recall questions on W = Fs cos θ and why the cosine appears, the work done by a force at right angles and at 180° to the motion, resolving the force against thinking of the displacement, and worked problems on a dragged or pushed load.
What you need to know
Answer each one before you reveal it.
Explain the origin of the cos θ factor in W = Fx cos θ.
What is the work done by a force acting at right angles to the motion?
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation for the work done by a constant force acting at an angle to the direction of motion, defining every symbol and its unit.
Model answer
The equation is W = Fx cos θ, where W is the work done in J, F is the force in N, x is the distance moved in m and θ is the angle between the force and the direction of motion.
- 2
Explain the origin of the cos θ factor in W = Fx cos θ.
Model answer
Only the component of the force along the direction of motion, F cos θ, moves through the distance x, and the perpendicular component does no work at all.
- 3
What is the work done by a force acting at right angles to the motion?
Model answer
The work done is zero, because cos 90° is zero and the force has no component along the direction of travel.
- 4
A rope pulls a sledge 15.0 m along level ground with a force of 80.0 N at 25.0° above the horizontal. Calculate the work done by the rope.
Model answer
The work done is 80.0 × 15.0 × cos 25.0° = 1.09 × 10³ J. The horizontal component is 80.0 cos 25.0° = 72.5 N, and 72.5 × 15.0 = 1088 J confirms the result.
- 5
What value does cos θ take when a force directly opposes the motion, and what does this mean for the work done?
Model answer
The angle is 180° so cos θ is −1, and the work done by the force is negative, meaning energy is taken from the moving body.
- 6
A friction force of 45 N acts on a box pushed 8.0 m along a floor. Calculate the work done by friction.
Model answer
Friction acts at 180° to the motion, so the work done by it is 45 × 8.0 × cos 180° = −360 J. The negative sign shows that 360 J is removed from the box.
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