Potential Energy of an Object at a Height – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Define gravitational potential energy and derive the formula PE = mgh. Explain each step and the meaning of each symbol.

Answer:

Gravitational potential energy (PE) is the energy an object has because of its position in a gravitational field, usually above the ground. It equals the work done to lift the object from the
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level to that height.
To derive the formula, consider an object of mass m lifted vertically through a height h at constant speed. The force needed equals the object’s weight = mg, where g is the acceleration due to gravity.
Work done W = Force × Displacement = (mg) × h = mgh.
Since work done to raise the object is stored as potential energy, PE = mgh.
Units: mass in kg, g in m/s², height in m; PE in joules (J).
Key idea: PE depends on mass, height, and gravity; path does not matter, only vertical displacement.

Q2. A 10 kg box is lifted to 6 m. Calculate its potential energy using g = 9.8 m/s², and explain what this energy means in everyday terms.

Answer:

Using the formula PE = mgh: m = 10 kg, g = 9.8 m/s², h = 6 m.
PE = 10 × 9.8 × 6 = 588 J.
This 588 joules means that 588 joules of work were done against gravity to lift the box to 6 m, and that amount of energy is now stored in the box because of its position.
In everyday terms, if the box falls freely from that height, the stored potential energy can convert into kinetic energy and could, for example, do 588 J of work on impact (neglecting air resistance).
This energy corresponds to a small amount: 1 J is the energy to lift a small apple by about 10 cm, so 588 J is like lifting many apples—useful to visualize scale.

Q3. Explain why gravitational potential energy depends only on the vertical height and not on the path taken. Give a simple example to support your explanation.

Answer:

Gravitational force is a conservative force. For conservative forces, the work done depends only on the initial and final positions, not the path.
When lifting an object from ground level to height h, the change in potential energy equals the weight × vertical displacement (mgh). Any extra distance or curves taken do not change the vertical displacement, so they do not change PE.
Example: Carrying a book straight up a ladder to a shelf 2 m high or walking it up a long ramp that also rises 2 m results in the same increase in PE = mg × 2.
However, the ramp may require more total muscular effort because you have to apply force over a longer path; but the energy stored as gravitational potential is the same.
Key point: PE depends on vertical change; path affects the effort or power required but not the PE gained.

Q4. In the example where PE = 480 J, m = 12 kg, and g = 10 m/s², height was found to be 4 m. Explain how rounding g to 10 affects the answer, and when such approximations are acceptable.

Answer:

Given PE = mgh, height h = PE / (mg). Using g = 10 m/s² gives h = 480 / (12 × 10) = 4.0 m. If we used g = 9.8 m/s², h = 480 / (12 × 9.8) ≈ 4.0816 m.
The difference is about 0.08 m (8 cm), less than 2% error, which is small for many school problems. Approximating g = 10 m/s² simplifies calculations and is acceptable when high precision is not required, like quick estimates or theoretical examples.
However, for precise engineering tasks, scientific experiments, or where small differences matter, use g = 9.8 m/s² (or local g value).
Always state the value of g used and be aware that approximation introduces a small systematic error.

Q5. Design and explain an experiment (with expected calculations) to show how potential energy changes with height using a 1 kg mass. Include sources of error and how to reduce them.

Answer:

Procedure: Use a 1 kg mass, a measuring tape, and a spring scale to confirm weight. Measure heights 1 m, 2 m, 3 m. For each height compute PE = mgh with g = 9.8 m/s².
Expected calculations: PE(1 m) = 1 × 9.8 × 1 = 9.8 J; PE(2 m) = 19.6 J; PE(3 m) = 29.4 J. These values should increase linearly with height.
Record data in a table and plot PE vs height. The graph should be a straight line through the origin with slope mg (≈9.8 N).
Sources of error: imprecise height measurement, parallax reading of tape, air currents, inaccurate mass or spring scale.
Reduce errors by using a stable stand, measuring tape fixed vertically, repeating measurements and averaging, and using a calibrated mass.
This experiment demonstrates the direct, linear relationship between height and gravitational potential energy.

High Complexity (Analytical & Scenario-Based)

Q6. A 2 kg object is held at 10 m and then released. Using energy conservation, calculate its speed just before hitting the ground ignoring air resistance. Explain each step and the principle used.

Answer:

Principle: Conservation of mechanical energy—if no non-conservative forces (like air resistance) do significant work, the sum of PE and KE remains constant. Initially the object has PE = mgh = 2 × 9.8 × 10 = 196 J and KE = 0 (released from rest). Just before impact all PE converts into kinetic energy (KE).
Set KE_final = PE_initial → (1/2) m v² = 196. So v² = (2 × 196) / 2 = 196. Therefore v = √196 = 14 m/s.
Steps: compute initial PE, equate to final KE, solve for v.
Interpretation: The speed 14 m/s is obtained purely from gravitational potential energy converting to kinetic energy. If air resistance were present, final speed would be lower because some energy would be lost as heat and air drag work.

Q7. Two objects A (4 kg) and B (2 kg) are placed on shelves at heights 2 m and 4 m respectively. Which has greater potential energy? If both drop, which will reach the ground first if released simultaneously? Analyze and explain.

Answer:

Compute potential energies: PE_A = mgh = 4 × 9.8 × 2 = 78.4 J. PE_B = 2 × 9.8 × 4 = 78.4 J. Both have the same potential energy despite different masses and heights because m×h is equal.
If both are released simultaneously from rest and air resistance is negligible, they will both take the same time to fall because the time of free fall depends only on height (t = √(2h/g)), not mass. However, since heights differ, their fall times differ: object A falls from 2 m, t_A = √(4/9.8) ≈ 0.64 s; object B from 4 m, t_B = √(8/9.8) ≈ 0.90 s. So A reaches first because it has a shorter height.
Summary: They have equal PE, but fall times depend on height, not mass. Equal PE does not imply equal fall time.

Q8. An object is lifted up an inclined plane to the same vertical height as when lifted vertically. Compare the work done by the lifting force in both cases and explain where energy differences (if any) appear.

Answer:

Work required to raise an object by a vertical height h against gravity equals mgh, independent of the path, so the minimum work done against gravity is the same for both vertical lift and incline. If lifting at constant speed, the net work by applied force equals increase in PE = mgh.
However, along an inclined plane, the applied force is smaller but acts over a longer distance. Total work done by the applied force against gravity still equals mgh.
Differences arise if there is friction on the plane. With friction, additional work is done to overcome frictional force (f × distance along plane), so total work by the person is mgh + work against friction, larger than vertical ideal case.
In practice, inclines are used to reduce required force at the cost of longer distance or possibly extra work if friction exists.

Q9. A satellite maintenance robot on the Moon lifts a 5 kg tool to a shelf 2 m above the lunar surface. Given lunar g ≈ 1.6 m/s², calculate the PE gained and explain how this would differ on Earth. Why is the energy smaller on the Moon?

Answer:

On the Moon: PE = mgh = 5 × 1.6 × 2 = 16 J. So the tool gains 16 joules of gravitational potential energy when lifted 2 m.
On Earth (g ≈ 9.8 m/s²): PE = 5 × 9.8 × 2 = 98 J. So the same lift requires much more energy on Earth.
The energy is smaller on the Moon because gravity is weaker there (about 1/6 of Earth’s). Since PE depends directly on g, a smaller g means less work is required to raise the tool to the same height.
Practical implication: Astronauts can lift heavy objects more easily on the Moon, but the mass (inertia) is unchanged, so changing motion still requires force. Energy comparison shows how gravitational field strength directly affects mgh.

Q10. A 3 kg mass is raised from 1 m to 5 m height. During lifting, a constant frictional force of 5 N acts along the vertical path due to a rough rope. Calculate the increase in the object’s PE and the additional work done against friction. Explain total work done by the person.

Answer:

Increase in height Δh = 5 − 1 = 4 m. Increase in PE = m g Δh = 3 × 9.8 × 4 = 117.6 J. This is the energy stored as gravitational potential.
Frictional force is 5 N acting along the path. Work done against friction = friction × displacement = 5 × 4 = 20 J. This energy is dissipated as heat in the rope.
Total work done by the person = increase in PE + work against friction = 117.6 + 20 = 137.6 J.
Explanation: The person supplies energy to both store potential energy in the mass and overcome non-conservative frictional forces. If no friction existed, person’s work would equal only 117.6 J. Here, friction increases the required energy and converts part of supplied work into heat.