Kinetic and Potential Energy Visualiser

Energy is the central concept of physics. Everything that happens — a ball rolling down a hill, a rocket launching, an atom emitting light — involves energy changing from one form to another. The two most fundamental mechanical forms are kinetic energy (energy of motion) and potential energy (energy of position).

The law of conservation of energy says the total mechanical energy stays constant in the absence of non-conservative forces like friction. Watch it in action as a mass descends a ramp.

Total energy

490.5 J

Speed at bottom

14.0 m/s

KE at bottom

490.5 J

PE at top

490.5 J

5 kg
10 m
0 m/s

The Formulas

Kinetic Energy

KE=12mv2KE = \frac{1}{2}mv^2

Kinetic energy depends on mass and the square of velocity. Doubling the speed quadruples the kinetic energy.

Gravitational Potential Energy

PE=mghPE = mgh

Potential energy depends on mass, gravitational acceleration g (9.81 m/s²), and height above the reference level.

Conservation of Mechanical Energy

In the absence of friction:

KE1+PE1=KE2+PE2KE_1 + PE_1 = KE_2 + PE_2

12mv12+mgh1=12mv22+mgh2\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2

SymbolQuantityUnit
mMasskg
vSpeedm/s
gGravitational accelerationm/s² (9.81 on Earth)
hHeightm
KEKinetic energyJ (joule)
PEPotential energyJ (joule)

Worked Examples

Worked Example

Example 1 — Ball dropped from rest

A 2 kg ball is dropped from a height of 20 m. What is its speed just before hitting the ground? (Ignore air resistance.)

Using conservation of energy:

At the top: all energy is PE = mgh = 2 × 9.81 × 20 = 392.4 J, KE = 0

At the bottom: all energy is KE, PE = 0

12mv2=mghv=2gh=2×9.81×2019.8 m/s\frac{1}{2}mv^2 = mgh \quad \Rightarrow \quad v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} \approx 19.8 \text{ m/s}

Note: the mass cancels — all objects (regardless of mass) reach the same speed when dropped from the same height.

Worked Example

Example 2 — Roller coaster minimum speed

A roller coaster cart starts from rest at a height of 40 m. What is the minimum height a loop of radius 8 m can be at for the cart to complete the loop?

At the top of the loop, the cart needs a minimum centripetal acceleration of g:

vmin2=gR=9.81×8=78.5 m2/s2v_{min}^2 = gR = 9.81 \times 8 = 78.5 \text{ m}^2\text{/s}^2

Using energy conservation from start height HH to loop top height hloop+2Rh_{loop} + 2R:

mgH=mg(hloop+2R)+12mvmin2mgH = mg(h_{loop} + 2R) + \frac{1}{2}mv_{min}^2

If the loop sits at ground level (hloop=0h_{loop} = 0):

hloop=H2Rvmin22g=40164=20 mh_{loop} = H - 2R - \frac{v_{min}^2}{2g} = 40 - 16 - 4 = 20 \text{ m}

The loop top is at 16 m. Since 40 m > 16 m, the cart completes the loop with speed to spare.


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