Work, Energy, and Simple Machines · Lesson 7 of 13
Gravitational Potential Energy
“Height gives an object stored energy without making it look any more impressive.”
• Relate gravitational potential energy to the Earth-object system. • Use an investigation to connect height with stored energy. • Derive gravitational potential energy near Earth’s surface. • Calculate potential-energy changes using mass, gravity and vertical height. • Explain reference level and path independence.
Drop a heavy ball onto loose sand from a small height and it makes a depression. Drop the same ball from a greater height and the depression becomes deeper. The ball did not become heavier, but lifting it farther from Earth required more work. The raised Earth-ball system therefore stored more energy and could do more work on the sand during impact.
Activity: Let Us Investigate
Fill a broad container with loose sand and smooth its surface. Hold a heavy ball about one metre above the sand and release it without throwing. Observe the depression. Smooth a fresh region, raise the same ball to about two metres and release it again. Repeat if needed at separate positions so the depressions do not overlap. Compare their depths.
The ball released from the greater height generally produces the deeper depression. Raising it farther required more work against gravity. When the ball falls, the stored energy becomes kinetic energy and is then transferred to the sand as the ball pushes grains aside. The result provides observable evidence that gravitational potential energy increases with height.
Use a stable container on the floor, keep feet and hands away from the falling ball, and choose a ball that cannot shatter or rebound dangerously.
Choosing a Reference Level
A height is measured from a chosen zero level. In a simple situation, the ground may be assigned zero potential energy. A tabletop or the lowest point of a pendulum can also serve as the reference when that choice makes the comparison clearer. Potential-energy changes are physically meaningful; the numerical zero is selected for convenience.
Deriving the Expression
Consider an object of mass m raised slowly from the chosen zero level to vertical height h. Near Earth’s surface, its weight is mg. To raise it gradually without changing its kinetic energy, the applied upward force has magnitude mg. The force and displacement are both upward.
The work-energy theorem states that this work appears as a change in the system’s energy. If potential energy is chosen to be zero at the starting level, the potential energy at height h is:
| Quantity | Effect on Potential Energy |
|---|---|
| Mass m | Doubling mass doubles U |
| Gravitational acceleration g | Larger g gives larger U for the same mass and height |
| Vertical height h | Doubling height doubles U |
| Path shape | Does not affect the gain when initial and final heights are unchanged |
Problem
A 200 g cricket ball reaches a height of 10 m. Find its gravitational potential energy using g = 10 m s⁻² and ground as the zero level.
- 1.Convert mass to kilograms: 200 g = 0.2 kg.
- 2.Write U = mgh.
- 3.Substitute U = 0.2 kg × 10 m s⁻² × 10 m.
- 4.Multiply the values: U = 20 kg m² s⁻².
- 5.Therefore, the gravitational potential energy is 20 J.
Problem
A 50 kg student reaches a height of 72.5 m, first by elevator and later by stairs. Find the potential-energy gain in each case using g = 10 m s⁻².
- 1.The initial and final heights are the same in both journeys.
- 2.Use ΔU = mgh.
- 3.ΔU = 50 kg × 10 m s⁻² × 72.5 m.
- 4.ΔU = 36250 J.
- 5.The gain is 36250 J by elevator and 36250 J by stairs.
- 6.Gravitational potential-energy change depends on vertical height, not on the path taken.
Problem
An object moves horizontally at constant height and is then raised vertically. How does its gravitational potential energy change?
- 1.During horizontal motion, its height relative to the chosen reference stays unchanged.
- 2.Since m, g and h are unchanged, its gravitational potential energy remains unchanged.
- 3.During vertical lifting, h increases.
- 4.Its gravitational potential energy therefore increases by mg multiplied by the vertical rise.
Range of the Expression
The expression U = mgh is used for heights near Earth’s surface where gravitational acceleration can be treated as constant. Far from Earth, the value of g changes noticeably and a different expression is required. Within the situations studied here, treating g as constant is appropriate.
Comparing Changes Without Full Calculation
Because U is directly proportional to both mass and height, many comparisons can be made before inserting values. If one object has twice the mass but rises through half the height, its potential-energy gain is the same as that of the original object. If both mass and height double, the gain becomes four times as large. Such proportional reasoning provides a quick check on numerical answers.
Problem
Object A has mass m and is raised through height h. Object B has mass 3m and is raised through height 2h. Compare their gravitational potential-energy gains.
- 1.For object A, ΔUA = mgh.
- 2.For object B, ΔUB = (3m)g(2h).
- 3.Multiply the factors: ΔUB = 6mgh.
- 4.Therefore, object B gains six times as much gravitational potential energy as object A.
In U = mgh, h is the vertical difference between the initial and final levels. It is not the length of a staircase, slide or sloping ramp.
Quiz
Which statement is a key takeaway from this lesson?
Which additional statement is also a key takeaway from this lesson?
Which further statement is also a key takeaway from this lesson?
Which another statement is also a key takeaway from this lesson?
Which final statement is also a key takeaway from this lesson?
Practice Problems
- Find the potential energy of a 3 kg object held 5 m above the chosen zero level when g = 10 m s⁻².
- Two objects of masses 2 kg and 6 kg are raised to the same height. Compare their potential energies.
- A box reaches the same platform by a short steep ramp and a long gentle ramp. Compare its potential-energy gain.
- Explain how the ball-and-sand investigation connects work, height and energy.
Key Takeaways
• Gravitational potential energy belongs to the interacting Earth-object system. • Near Earth’s surface, U = mgh. • It increases with mass and vertical height. • Its change is independent of the path between two fixed heights. • A reference level must be chosen before assigning a potential-energy value.