5.2 Gravitational potential energy and kinetic energy
- Syllabus
- 9702–2028–2029
- Topic
- 5.2
- Level
- AS
Learning objectives
5.2.1W = Fs, the formula ∆EP = mg∆h for gravitational potential energy changes in• derive, using W = Fs, the formula ∆EP = mg∆h for gravitational potential energy changes in a uniform gravitational field5.2.2The formula ∆EP = mg∆h for gravitational potential energy changes in a• recall and use the formula ∆EP = mg∆h for gravitational potential energy changes in a uniform gravitational field5.2.3The equations of motion, the formula for kinetic energy EK = 1/2 mv2• derive, using the equations of motion, the formula for kinetic energy EK = 1/2 mv25.2.4EK = 1/2 mv2• recall and use EK = 1/2 mv2
Derive gravitational potential energy change from work
Raise a mass m slowly through vertical height Δh in a uniform gravitational field, so the upward applied force equals its weight mg and the displacement is along that force.
Work done W = Fs. Substitute F = mg and s = Δh: W = mgΔh. This work transfers energy to the gravitational potential store, so ΔEP = W = mgΔh.
ΔEP=mgΔh
The derivation uses vertical height change, not path length, and assumes g is uniform. Raising gives positive ΔEP; lowering gives negative ΔEP when Δh is signed.
Use ∆E_P=mg∆h for a uniform gravitational field near Earth
In a uniform field, the change in gravitational potential energy is ∆E_P=mg∆h, with g treated as constant over the height interval.
Define the reference level and keep the sign of ∆h consistent. Only differences in potential energy affect energy conservation calculations.
A 0.50 kg mass lowered 4.0 m has ∆E_P=−19.6 J relative to its starting level; that energy can become kinetic or thermal.
Zero potential at the floor is a choice, not a physical claim that the object has no energy anywhere else.
Derive kinetic energy from resultant work and motion
Let a constant resultant force F accelerate a constant mass m through displacement s, changing its speed from u to v. The resultant work is W = Fs and F = ma, so W = mas.
From v² = u² + 2as, as = (v² − u²)/2. Substitute into W = mas: W = ½m(v² − u²) = ½mv² − ½mu².
Resultant work equals the change in kinetic energy, so EK = ½mv² relative to rest, and ΔEK = ½mv² − ½mu² for a speed change.
Kinetic energy uses speed squared and is scalar. The derivation assumes constant mass; signs of velocity disappear only after the vector dynamics have established the speed change.
Kinetic energy is the energy of motion, E_K=½mv²
The kinetic-energy store of a mass m moving at speed v is E_K=½mv². It is a scalar and is never negative.
Use speed magnitude and consistent units; doubling speed quadruples kinetic energy, while doubling mass doubles it.
A 4.0 kg trolley moving at 3.0 m s⁻¹ has kinetic energy 18 J.
Kinetic energy does not carry the direction sign of momentum, and stopping does not destroy it—it transfers it to other stores.