Kinetic potential and mechanical energy

Energy is the capacity to do work: a body moving with speed $v$ has kinetic energy $E_k = \frac{mv^2}{2}$, while a body of mass $m$ raised to height $h$ near the Earth's surface has gravitational potential energy $E_p = mgh$. Both are scalars measured in joules — the same unit as work $A = Fs\cos\alpha$ — so they can be added directly, and power $P = A/\Delta t$ only tells how fast this energy is converted. Their sum is the total mechanical energy of the body, $E = E_k + E_p$. When only conservative forces act (gravity, an ideal spring), the work they do is fully returned, so $E = const$: whatever potential energy is lost comes back as kinetic energy. If friction or air drag acts, mechanical energy is not conserved; its change equals the work of the non-conservative forces, $\Delta E = W_{nc} < 0$, and the missing part becomes internal energy — exactly what the efficiency of a real machine accounts for. A bar chart of $E_k$, $E_p$ and $E$ against height makes this balance visible at a glance.

A body of mass $m = 2{,}0$ kg moves with speed $v = 3{,}0$ m/s at a height $h = 5{,}0$ m above the ground. Taking $g = 9{,}8$ m/s², find its kinetic, potential, and total mechanical energy.

A sled slides from rest down a frictionless hill of height $h = 3{,}2$ m. (a) What is its speed at the foot of the hill? (b) At what height is its kinetic energy equal to its potential energy? Take $g = 9{,}8$ m/s².

A cart of mass $m = 0{,}60$ kg rolls from rest down an inclined rail of length $s = 6{,}0$ m from a height $H = 1{,}5$ m. At the bottom of the rail its speed is $v = 4{,}0$ m/s. Find the work of the friction force, the average friction force, and the fraction of mechanical energy that survives to the end of the descent. Take $g = 9{,}8$ m/s².

Simulation instead of a real slope. Do all the measuring in the interactive PhET model Energy Skate Park: do not try to repeat the experiment on a real skateboard or an inclined surface.

1. Switch to the energy mode, turn friction off, and place the skater at height $h = 2{,}0$ m. First predict the speed at the lowest point with $v = \sqrt{2gh}$, then switch on the speedometer and compare the prediction with the reading.
2. Fill in a table for three heights ($2{,}0$, $1{,}5$, $1{,}0$ m): $h$, the measured $v$, and the computed $E_k = \frac{mv^2}{2}$, $E_p = mgh$ and $E = E_k + E_p$. Check that $E$ is constant to within about 5 %.
3. Switch on the friction slider and repeat: record $E$ at the start and at the end of one descent. Verify that the difference equals the work of the friction force and that the total bar height now drops. State the conclusion in one sentence: when $E$ is conserved and why it stops being conserved.

Self-check: on the chart, find the point where the $E_k$ and $E_p$ bars are equal and compare its height with $h/2$.

Show that in frictionless motion the sum $E_k + E_p$ stays constant, while with friction it decreases by exactly the work of the friction force.

On the left, a sketch of the hill with marked points A (top), B (middle, where $E_k = E_p$), and C (bottom). On the right, below it, an energy bar chart: for each point two bars $E_k$ and $E_p$ that always add up to the same height — the level $E = mgh_A$ drawn as a dashed line. Below that, a third set of bars for the case with friction: the total height is lower than the dashed level by $|W_{fr}|$, and the missing part is shown as a grey block labelled 'internal energy'.

A skateboarder of total mass $55$ kg rides down a ramp $2.0$ m high, starting from rest at the top. At the bottom his speed is $v=\sqrt{2gh}=\sqrt{2\cdot 9.8\cdot 2.0}\approx 6.3$ m/s and his kinetic energy is $\frac{1}{2}mv^2=mgh=55\cdot 9.8\cdot 2.0\approx 1.1\cdot 10^3$ J.

This uses conservation of total mechanical energy $E=E_k+E_p$: at the top $E_p=mgh\approx 1.1$ kJ and $E_k=0$; at the bottom $E_p=0$ and all of it has become $E_k$. If friction on the ramp did noticeable work, the two totals would not match, and their difference would equal the work of friction.

A car of mass $1200$ kg travels at $90$ km/h $=25$ m/s. Its kinetic energy is $E_k=\frac{1}{2}mv^2=\frac{1}{2}\cdot 1200\cdot 25^2=3.75\cdot 10^5$ J $=375$ kJ. With a friction coefficient $\mu=0.7$ the braking force is $F=\mu mg\approx 8.2\cdot 10^3$ N, so the stopping distance is $s=E_k/F\approx 46$ m.

This is the work-kinetic-energy statement: the work of friction $W=-Fs$ equals the change in kinetic energy $\Delta E_k=0-375$ kJ. The dependence on speed is quadratic: at $180$ km/h the energy is four times larger and the braking distance grows to $\approx 183$ m.

A phone of mass $0.20$ kg slides off a table $1.2$ m high. By the moment it hits the floor its potential energy $mgh=0.20\cdot 9.8\cdot 1.2\approx 2.4$ J has turned into kinetic energy, so the impact speed is $v=\sqrt{2gh}\approx 4.9$ m/s. The impact is inelastic: mechanical energy turns into internal energy — heating and deformation of the case and the glass.

This shows where conservation of mechanical energy holds in its pure form and where it does not: while the phone is in the air $E_k+E_p$ stays constant, but at the impact with the floor mechanical energy turns into other forms, even though the total energy is conserved.