potential energy and conservation of mechanical energy

From the study of work and kinetic energy, the work of the net force equals the change in kinetic energy: $W=\Delta K$.

For conservative forces, it is convenient to write work as the decrease in potential energy: $W=-\Delta U$.

In a uniform gravitational field $U_g=mgh$, and for an elastic spring $U_s=\frac12 kx^2$, where $x$ is the deformation from the unstretched state.

The total mechanical energy $E=K+U$ is conserved if the work done by nonconservative forces is zero: $E_1=E_2$.

For example, $\frac12mv_1^2+mgh_1+\frac12kx_1^2=\frac12mv_2^2+mgh_2+\frac12kx_2^2$.

If friction or an external force acts, use the balance $W_{\text{noncons}}=\Delta E=E_2-E_1$.

A ball of mass $m=0.50$ kg is released from rest from height $h=2.0$ m. Ignore air resistance. Find the ball's speed just before it hits the ground.

A block of mass $m=0.20$ kg starts from rest and slides without friction from height $h=1.5$ m, then compresses a spring of stiffness $k=250$ N/m on a smooth horizontal section. Find the maximum spring compression $x$.

A skier of mass $m=60$ kg starts from rest from height $h=20$ m. At the bottom, the skier's speed is $v=15$ m/s. Find the work done by friction and drag along the entire descent.

Use the safe PhET simulation “Energy Skate Park: Basics” or an equivalent. 1) Turn friction off, set an initial height, and record $h$ and $v$ at two points; compute $K$, $U$, $E=K+U$ and verify $E_1=E_2$. 2) Turn friction on and compare $E_1$ with $E_2$; find $W_{\text{noncons}}=E_2-E_1$. Do not perform real jumps, descents, or measurements with moving loads; perform all observations in the simulation.

Show the exchange between kinetic and potential energy and the constancy of total mechanical energy when there is no friction.

A sketch of a skate track or pendulum with three marked positions and energy bar charts: $K$, $U$, $E=K+U$ at each position; with friction on, the final $E$ bar is lower than the initial one.

A $70\ \mathrm{kg}$ ski jumper starts almost from rest at a height of $40\ \mathrm{m}$ above the takeoff point. Ignoring friction and air drag, the gravitational potential energy is $U=mgh=70\cdot9.8\cdot40=27\,440\ \mathrm{J}$. At takeoff this energy turns into kinetic energy: $K=\frac{1}{2}mv^2$, so $v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot40}\approx28\ \mathrm{m/s}\approx100\ \mathrm{km/h}$.

Conservation of mechanical energy applies here: while only gravity does work, the sum $E=K+U$ stays constant. Gravitational potential energy $U=mgh$ converts into kinetic energy $K=\frac{1}{2}mv^2$; mass cancels when the speed is found.

A $1200\ \mathrm{kg}$ car starts down a hill of height $25\ \mathrm{m}$ at a speed of $5\ \mathrm{m/s}$. Ignoring rolling resistance, the speed at the bottom follows from $\frac{1}{2}mv_i^2+mgh=\frac{1}{2}mv_f^2$: $v_f=\sqrt{v_i^2+2gh}=\sqrt{25+2\cdot9.8\cdot25}\approx22.7\ \mathrm{m/s}\approx82\ \mathrm{km/h}$.

This is a conservation-of-mechanical-energy problem with initial kinetic energy. The term $mgh$ gives the potential-energy store that is added to the initial $\frac{1}{2}mv_i^2$ and becomes the final $\frac{1}{2}mv_f^2$.

A $0.18\ \mathrm{kg}$ smartphone falls from a table of height $0.80\ \mathrm{m}$. Just before hitting the floor, its potential energy relative to the floor was $U=mgh=0.18\cdot9.8\cdot0.80=1.41\ \mathrm{J}$, and the speed ignoring air drag is $v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot0.80}=3.96\ \mathrm{m/s}$.

Before impact, mechanical energy is conserved: $U=mgh$ converts into $K=\frac{1}{2}mv^2$. At the impact, non-conservative forces act—case deformation, sound, and heating—so mechanical energy is no longer conserved and turns into internal energy.