Energy on a ramp

Where does energy go when a block descends and gains speed?

Block on a ramp with an energy balance in three barsH = 4 mh = 4 m · v = 0 m/s30°Bottom: h = 0 mEnergy (J)Potential78,48Kinetic0Thermal0Sum = 78,48 Jt = 0 s / 1,81 s · The bars share a scale of 78,48 J.
Animation at 0,5× · simplified model
Instant: 0 s

One comparison at a time

Experiment with a guide

Review the foundation lesson →

Recognise the transformation between potential, kinetic and thermal energy and check the energy balance.

Comparison 1 of 3

Descend without dissipation

Use 2 kg at 4 m above the bottom and a loss of 0%. Advance the animation to the bottom. All the initial potential energy can become kinetic energy.

Initially there is 78,48 J of potential energy; at the bottom there is 78,48 J of kinetic energy and a speed of about 8,86 m/s.

How to read this drawing

The block starts at the top of a 30° ramp. The vertical line marks the height still to descend. The three bars use the same scale in joules: lilac is potential energy, green is kinetic and amber is thermal. Zero height is at the bottom. The ramp adapts to the initial height; compare the indicated heights and values in J.

From drawing to graph

What changes? Let’s compare.

A line shows the relationship between quantities. The point marks your choice or the instant in the animation. First read the axis names and units.

Instant: 0 s

Read the graph

Current: solid line. Saved: dashed line. Slide to see the details.

One experiment at a time

Save. Change. Compare.

Save the current values, change one quantity and observe both lines on the same graph. The saved line is dashed. The attempt stays in this browser.

Compare the chosen values

What if… Compare the same descent without losses and with 20% losses. Then double the mass and compare speed and energies.

What is happening?

As it descends, the block loses gravitational potential energy and gains kinetic energy. With losses, part becomes thermal energy.

The initial height stores potential energy mgh. The block starts at rest on a 30° ramp. As it descends, potential energy decreases. Without losses, this entire decrease becomes kinetic energy. The chosen loss represents the fraction of gravity’s work converted into thermal energy along the ramp. The sum of all three forms of energy stays the same.

Eₚ = mgh · E꜀ = ½mv² · Eₚ + E꜀ + Eₜ = initial E

m — mass (kg) · h — height (m) · v — speed (m/s) · g = 9,81 m/s² · E — energy (J)

The model and its limits

Block starting at rest, fixed straight 30° ramp and constant gravity. The dissipative force is constant and equals the chosen fraction of the weight component along the ramp. The animation ends at the bottom; it does not simulate motion after the ramp. The drawing adapts to height; the bars share the initial-energy scale.

Put your idea to the test

3 questions
1A 2 kg block starts at a height of 4 m. With g = 9,81 m/s², what is its initial potential energy?

2With a 20% loss along the ramp, what part of the initial energy is kinetic at the bottom?

3On the same ramp, with the same height and loss percentage, doubling the mass makes the final speed…