Motion in a Gravitational Field: projectiles, energy and drag
Key idea: Use one uniform gravitational-field model consistently across force, component-motion and energy descriptions, then identify how drag changes the motion.
Before you start: Kinematics GraphsEnergy & Fields
By the end, you can
- Describe weight as the gravitational force on a mass and use W = mg in a uniform field.
- Explain projectile motion as uniform velocity in one direction and uniform perpendicular acceleration in the other.
- Derive ΔEₚ = mgΔh from work done and apply it with a consistent height change and sign.
- Describe falling with air resistance using resultant force, energy transfers and terminal velocity.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first force, component or energy statement you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Uniform-field potential energy 5(c)–(d)
Question 1
Starting from work done by a constant force, derive the gravitational potential-energy change for lifting a mass m through a vertical height Δh in a uniform field.
Check the model response
For a slow lift, the upward applied force has magnitude mg. Work transferred to the gravitational potential-energy store is force times displacement in its direction: W = (mg)Δh. Therefore ΔEₚ = mgΔh.
repair
2. Repair the five common breaks
Use only the repair matching an error, then rebuild the force, component or energy model.
Uniform-field potential energy 5(c)–(d)
Check this idea
Misconception: The equation mgh gives one absolute potential energy.
Repair: In a uniform field, ΔEₚ = mgΔh gives a change relative to chosen heights. Only the height difference matters; raising gives positive Δh and lowering gives negative Δh.
Check this idea
Misconception: Gravitational potential-energy change depends on the path taken between two heights.
Repair: Gravity is conservative. In the uniform-field model, ΔEₚ depends only on the vertical height change: mgΔh, not the route length.
worked example
3. Follow three worked models
Follow how each solution declares axes, system boundaries and the uniform-field assumption.
Uniform-field potential energy 5(c)–(d)
Model 1
A 2.5 kg load is raised slowly through 6.0 m where g = 9.8 N kg⁻¹. Derive the relationship used, then find its potential-energy increase and work done by gravity.
Check the model response
The lifting force is mg, so transferred work W = force × upward displacement = mgΔh = ΔEₚ. Thus ΔEₚ = 2.5(9.8)(6.0) = +147 J. Gravity acts opposite the displacement, so its work is −147 J.
guided practice
4. Guided practice
Use each hint only to choose the force, component equation or energy-transfer direction.
Uniform-field potential energy 5(c)–(d)
Question 1
A 5.0 kg object is lowered by 3.0 m in a uniform field where g = 9.8 N kg⁻¹. Find ΔEₚ and the work done by gravity.
Hint: Take upward height change as positive.
Check the model response
Δh = −3.0 m, so ΔEₚ = 5.0(9.8)(−3.0) = −147 J. Work done by gravity is −ΔEₚ = +147 J.
independent practice
5. Independent practice
Solve without repair notes and state every sign convention and modelling assumption.
Uniform-field potential energy 5(c)–(d)
Question 1
Explain why ΔEₚ = mgΔh follows from work done, including the assumptions, sign of Δh and why the chosen zero height does not affect the answer.
Check the model response
In a uniform field, weight mg is constant. Moving slowly through vertical displacement Δh requires opposite applied force mg, so transferred work is mgΔh and equals the potential-energy change. Upward Δh is positive and downward negative. Changing the zero adds the same constant to both endpoint energies, so their difference is unchanged.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Uniform-field potential energy 5(c)–(d)
Question 1
A 1200 kg car climbs through a vertical height of 25 m in a uniform field where g = 9.8 N kg⁻¹. Derive the potential-energy relationship, then find ΔEₚ and gravity’s work.
Check the model response
Lifting against constant weight mg through Δh transfers work mgΔh into gravitational potential energy, so ΔEₚ = mgΔh. Here ΔEₚ = 1200(9.8)(25) = +2.94 × 10⁵ J and work by gravity is −2.94 × 10⁵ J.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Uniform-field potential energy 5(c)–(d)
Question 1
A 4.0 kg package descends 7.5 m where g = 9.8 N kg⁻¹. Find its gravitational potential-energy change and the work done by the gravitational field.
Check the model response
Δh = −7.5 m, so ΔEₚ = 4.0(9.8)(−7.5) = −294 J. The gravitational field does +294 J of work.