Physics: Forces and Energy
Forces and energy is the physics topic in 13+ Common Entrance Science covering contact and non-contact forces, calculating resultant force, moments, pressure and density, and finding gravitational potential energy, kinetic energy and work done using standard equations. Pupils must be able to substitute values into formulae correctly and give answers with the right units.
Before you start
Make sure you're comfortable with these topics first:
Method
- Identify whether a force in the question is a contact force (for example friction, air resistance, normal contact) or a non-contact force (for example gravity, magnetism, static electricity).
- Write down the correct equation for the quantity you need before substituting numbers in, and always include units.
- For resultant force, moments and pressure questions, substitute the given values directly into the equation and calculate step by step.
- For 'explain' questions about motion, such as terminal velocity or a balanced beam, describe how the forces change and compare their sizes.
- Check that your final answer has the correct unit (N, m/s^2, J, Pa, kg/m^3) and is a sensible size for the situation described.
- For two-part energy questions, remember conservation of energy: gravitational potential energy lost is transferred to kinetic energy if there is no friction or air resistance.
Worked example
A go-kart of mass 80 kg travels at a speed of 6 m/s. Use the equation: kinetic energy (KE) = 0.5 x mass x speed^2. (a) Calculate the kinetic energy of the go-kart. (b) The go-kart slows to 3 m/s. Calculate its new kinetic energy.
- Substitute into KE = 0.5 x mass x speed^2: KE = 0.5 x 80 x 6^2.
- Calculate 6^2 = 36, then 0.5 x 80 x 36 = 1440.
- So the go-kart's kinetic energy at 6 m/s is 1440 J.
- For the new speed, substitute again: KE = 0.5 x 80 x 3^2 = 0.5 x 80 x 9 = 360.
- Final answer: KE at 6 m/s = 1440 J; KE at 3 m/s = 360 J.
Practice questions
Try each question, then tap to reveal the answer.
Exam-style questions
Written in the style of a 13+ Common Entrance exam paper, with a full mark scheme.
A uniform beam is pivoted at its centre. A force of 80 N pushes down on the beam at a distance of 0.4 m from the pivot on the left-hand side. (a) Calculate the force, F, that must be applied at a distance of 0.5 m from the pivot on the right-hand side for the beam to balance. Use moment = force x distance from pivot. (b) State the rule used to answer part (a).
A skydiver jumps from a plane and falls before opening her parachute. (a) State which statement correctly describes the forces acting on her just after she jumps, before she reaches terminal velocity: weight and air resistance are equal so she falls at a constant speed; OR weight is greater than air resistance so she accelerates downwards; OR air resistance is greater than weight so she accelerates upwards. (b) Explain, in terms of forces, why she eventually reaches a constant (terminal) velocity as she continues to fall.
Priya and Sam sit on a seesaw pivoted at its centre. Priya has a weight of 350 N and sits 1.6 m from the pivot. Sam has a weight of 400 N. (a) Calculate the distance from the pivot at which Sam must sit for the seesaw to balance. (b) Sam then moves to sit 1.3 m from the pivot, without changing his weight. State, with a calculation, what will happen to the seesaw.
Free printable worksheet
Want more practice on paper? Download the physics: forces and energy worksheet pack - 15 pages of exam-style questions with a full mark scheme. No sign-up, no email wall - just the PDF, free for personal and classroom use.
Next topics
Build a full practice pack.
This topic is one of hundreds in the library - pick the ones a student needs and generate a printable PDF in minutes.