Energy

Conservation of Energy

speed without forces or time

On a frictionless track potential energy trades cleanly into kinetic energy, so the speed at the bottom follows from the starting height alone.

mechanics Dropping 100 J from 5 m turns entirely into kinetic energy at the bottom, giving a clean 10 m/s with no forces or clock.

Example

On a frictionless track, potential energy trades cleanly into kinetic energy, so the speed at the bottom follows from the starting height alone — no forces, no clock. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start at the top with only potential energy

Release a 2 kilogram cart from rest at the top, 5 metres up, on a frictionless track. All its energy is potential: 2 times 10 times 5, which is 100 joules, and the kinetic energy is zero.

E=PE=2 kg ⋅ 10 m/s2 ⋅ 5 m=100 JE = PE = 2\ \text{kg} \,\cdot\, 10\ \text{m}/\text{s}^{2} \,\cdot\, 5\ \text{m} = \hl{100}\ \text{J}
At the top: all potential energyA cart at rest at the top of a sloped frictionless track.m

Partway down the buckets are both half full

Because the track is frictionless, no energy leaks away as heat, so the total stays 100 joules the whole way down. Halfway down, half the potential energy has already turned into kinetic: 50 joules of potential and 50 joules of kinetic, still adding to 100.

PE=50 J,KE=50 J,PE+KE=100 JPE = 50\ \text{J}, \quad KE = 50\ \text{J}, \quad PE + KE = \hl{100}\ \text{J}
Partway down: potential trades into kineticThe cart partway down the track, lower than the top and not yet at the bottom.m

At the bottom all of it is kinetic

At the bottom the height is zero, so the potential energy is zero and all 100 joules are kinetic. Setting one half times the mass times the speed squared equal to 100 gives the speed.

12 m v2=100  ⇒  v=10 m/s\tfrac{1}{2}\,m\,v^{2} = 100 \;\Rightarrow\; v = \hl{10}\ \text{m}/\text{s}
At the bottom: all kinetic energyThe cart at the bottom of the track, rolling onto flat ground with a velocity arrow showing its speed.mv

Potential plus kinetic keeps the same total

Read three positions as an energy ledger. Potential energy goes down, kinetic energy goes up, and the sum stays fixed.

PEKEE100 J0 J100 J60 J40 J100 J0 J100 J100 J\begin{array}{c|c|c}PE & KE & E \\ \hline 100\ \text{J} & 0\ \text{J} & 100\ \text{J} \\ 60\ \text{J} & 40\ \text{J} & 100\ \text{J} \\ 0\ \text{J} & 100\ \text{J} & 100\ \text{J}\end{array}

More drop height gives more bottom speed

For clean square cases, the bottom speed can be read exactly. More starting height gives more kinetic energy and therefore more speed at the bottom.

hPEv0 m0 J0 m/s5 m100 J10 m/s20 m400 J20 m/s\begin{array}{c|c|c}h & PE & v \\ \hline 0\ \text{m} & 0\ \text{J} & 0\ \text{m}/\text{s} \\ 5\ \text{m} & 100\ \text{J} & 10\ \text{m}/\text{s} \\ 20\ \text{m} & 400\ \text{J} & 20\ \text{m}/\text{s}\end{array}

Found without forces or time

Notice what we never used: no force, no acceleration, no time. Energy conservation jumped straight from the starting height to the final speed. That is the power of a conservation law — and because the mass is on both sides, it cancels, so every object reaches 10 metres per second from this height. It also did not matter that the track was straight: any frictionless path down from 5 metres, curved or steep, gives the same 10 metres per second, because only the height drop counts.

m g h=12 m v2  ⇒  v=2 g hm\,g\,h = \tfrac{1}{2}\,m\,v^{2} \;\Rightarrow\; v = \sqrt{2\,g\,h}