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)
Dynamics
Free body diagrams and connected bodies
- 3 things to know
- AS Unit 1
Overview
Recall questions on Free body diagrams and connected bodies.
What you need to know
Answer each one before you reveal it.
What forces belong on a free body diagram of an object, and what must be left off?
Why is the string in a connected-body problem usually taken to be light and inextensible?
Questions from this subtopic
Each one with the model answer underneath.
- 1
What forces belong on a free body diagram of an object, and what must be left off?
Model answer
Only the forces acting on that one object are shown, each with its direction. The forces that the object itself exerts on other bodies are always left off.
- 2
Why is the force a book exerts on a table left off the free body diagram of the book?
Model answer
That force acts on the table and not on the book, so it plays no part in the resultant force on the book or in the book's acceleration.
- 3
Name the four forces acting on an aircraft in steady level flight and state how they are related.
Model answer
The four forces are weight, lift, thrust and drag. In steady level flight the lift equals the weight and the thrust equals the drag.
- 4
A 70 kg passenger stands in a lift accelerating upwards at 1.5 m s⁻². Calculate the normal contact force from the floor, taking g as 9.81 m s⁻².
Model answer
The resultant upward force is N − mg = ma, so N = 70 × (9.81 + 1.5) = 792 N.
- 5
A 70 kg passenger stands in a lift accelerating downwards at 1.5 m s⁻². Calculate the normal contact force from the floor, taking g as 9.81 m s⁻².
Model answer
The resultant downward force is mg − N = ma, so N = 70 × (9.81 − 1.5) = 582 N.
- 6
Why does a passenger in a lift whose cable has broken feel weightless?
Model answer
The lift and the passenger fall with the same acceleration, so the floor no longer needs to push on the passenger at all and the contact force falls to zero. The passenger's weight is unchanged throughout.
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