Potential Energy Calculator

Potential energy is energy stored by an object's position or configuration. Near the Earth's surface, lifting a mass stores

$PE_g = mgh$

Symbols and SI units:

• $PE_g$ — gravitational potential energy, joules (J)
• $m$ — mass, kilograms (kg)
• $g$ — gravitational field strength, $9.81$ m/s² at the Earth's surface
• $h$ — height above the chosen reference level, metres (m)

Only the change in $PE$ has physical meaning, so you are free to put $h = 0$ wherever it is convenient — the floor, the table top, the ground. Different choices shift every value by the same constant and change no answer.

When it applies: $mgh$ assumes $g$ is constant, which holds while $h$ is small compared with the Earth's radius. For satellite-scale heights use $U = -GMm/r$ instead.

The assumption people forget: $h$ is the vertical rise only. Dragging a box $10$ m up a ramp that climbs $2$ m stores $mg(2\ \text{m})$, not $mg(10\ \text{m})$.

A stretched or compressed spring stores

$PE_e = \tfrac{1}{2}kx^2$

• $k$ — spring constant, newtons per metre (N/m)
• $x$ — extension or compression from the natural length, metres (m)

Because $x$ is squared, stretching and compressing by the same amount store the same energy, and doubling the stretch stores four times as much.

Conservation of mechanical energy ties potential energy to motion. With no friction or air resistance,

$PE_i + KE_i = PE_f + KE_f$

so a mass falling from rest through a height $h$ arrives at $v = \sqrt{2gh}$, independent of its mass.

When it applies: $\tfrac{1}{2}kx^2$ needs a spring obeying Hooke's law, $F = kx$, within its elastic limit. Conservation needs only conservative forces; friction dissipates energy as heat and breaks the balance.

The assumption people forget: $x$ is measured from the unstretched length, not from the floor or from any other datum.

• Using the distance travelled as $h$ — only the vertical component counts.
• Forgetting $g$ — a $5$ kg box raised $3$ m stores $147$ J, not $15$ J.
• Using weight in newtons as $m$ — divide by $g$ first to get kilograms.
• Measuring the spring extension from the wrong origin — it must be from the natural length.
• Halving after squaring in $\tfrac{1}{2}kx^2$, or squaring $k$ — only $x$ is squared.
• Leaving $x$ in centimetres — $12$ cm is $0.12$ m; forgetting this changes the answer by $10\,000$.
• Applying energy conservation with friction present — subtract the energy lost to friction, or the final speed comes out too high.
• Thinking a negative $PE$ is an error — below the reference level it simply is negative.