What if the mass is not given?

In many pure gravity problems, mass appears in every term (kinetic and gravitational potential energy) and cancels out algebraically, as shown in the example. If springs or non-conservative forces like friction are involved, mass typically does not cancel.

How do I handle a system with both a spring and gravity?

Include both potential energy terms in your initial and final states: $U = U_g + U_s = mgh + \frac{1}{2}kx^2$. Ensure the reference level for $h$ and the equilibrium point for $x$ are clearly defined.

Can gravitational potential energy be negative?

Yes. If an object moves below the chosen zero reference height ($h = 0$), its gravitational potential energy ($mgh$) will be negative. The choice of $h=0$ is arbitrary; only the change in potential energy affects the kinematics.

How to solve a conservation of energy problem

The conservation of mechanical energy principle states that in an isolated system with only conservative forces, the total mechanical energy remains constant. This means the sum of kinetic and potential energy at an initial state equals the sum at a final state.

Use this method when a system transitions between two states and non-conservative forces (like friction or applied forces) do zero net work, or when their work is explicitly accounted for in the generalized work-energy theorem.

The setup

Identify the system and verify that only conservative forces do work, or that you can account for non-conservative work. Define the initial state (1) and final state (2) of the system. Choose a reference level (zero point) for gravitational potential energy.

The steps

Write the general conservation of energy equation: $K_1 + U_1 + W_{nc} = K_2 + U_2$ , where $W_{nc}$ is the work done by non-conservative forces.
Expand the kinetic energy terms: $K = \frac{1}{2}mv^2$ .
Expand the potential energy terms. For gravity near Earth's surface, $U_g = mgh$ . For an ideal spring, $U_s = \frac{1}{2}kx^2$ .
Eliminate any terms that are zero based on your chosen reference frame and initial/final conditions.
Solve the resulting algebraic equation for the unknown variable.

Checking the result

Ensure the final answer has the correct units (e.g., meters per second for velocity, Joules for energy). Verify that the result makes physical sense; for instance, an object dropped from rest cannot have a final kinetic energy greater than its initial potential energy.

Common errors

Forgetting to include the work done by friction or other non-conservative forces when they are present.
Choosing a confusing or inconsistent zero reference level for potential energy.
Mixing up the initial and final states when plugging in values.
Failing to square velocity or displacement in the kinetic and spring potential energy formulas.

Worked example

A $2.0 ext{ kg}$ block is dropped from rest from a height of $5.0 ext{ m}$ above the ground. Air resistance is negligible. Calculate the speed of the block just before it hits the ground.

Define state 1 as the moment the block is dropped, and state 2 as the moment just before it hits the ground. Set the ground as $h = 0$ for gravitational potential energy.

Equation: $K_1 + U_1 = K_2 + U_2$

State 1: $v_1 = 0 \implies K_1 = 0$ $h_1 = 5.0 ext{ m} \implies U_1 = mgh_1$

State 2: $v_2 = ? \implies K_2 = \frac{1}{2}mv_2^2$ $h_2 = 0 \implies U_2 = 0$

Substitute into the equation: $0 + mgh_1 = \frac{1}{2}mv_2^2 + 0$

Solve for $v_2$ : $mgh_1 = \frac{1}{2}mv_2^2$ $gh_1 = \frac{1}{2}v_2^2$ $v_2 = \sqrt{2gh_1}$

Plug in values ( $g = 9.8 ext{ m/s}^2$ ): $v_2 = \sqrt{2(9.8)(5.0)}$ $v_2 = \sqrt{98}$ $v_2 \approx 9.9 ext{ m/s}$

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References: University Physics with Modern Physics, 14th Edition (Young and Freedman) · OpenStax University Physics Volume 1, Chapter 8: Potential Energy and Conservation of Energy · Fundamentals of Physics, 10th Edition (Halliday, Resnick, Walker)