Work-energy theorem and energy conservation

The work of the net force changes the kinetic energy of a body: $A_{\text{net}} = \Delta E_k = \frac{mv_2^2}{2} - \frac{mv_1^2}{2}$ — this is the work–energy theorem, valid also for variable forces if $A_{\text{net}}$ is understood as the work of the total force along the whole path.

If only conservative forces act (gravity, elasticity), their work is best written through the decrease of potential energy: $A_{\text{cons}} = -\Delta E_p$. Substituting this into the theorem gives $\Delta E_k + \Delta E_p = 0$, that is $E_k + E_p = \text{const}$.

The total mechanical energy $E = E_k + E_p$ is conserved when non-conservative forces (friction, air drag, a driving force) do no work; if they do work, the mechanical energy changes by exactly that work: $\Delta E = A_{\text{nc}}$, and friction usually decreases $E$.

A practical solution scheme: choose the initial and final states, write $E$ for each of them, add the work of non-conservative forces if needed, and solve the resulting equation for the required quantity.

Keep the prerequisites in mind: potential energy is defined only for conservative forces, and the height $h$ is measured from a freely chosen zero level.

A block of mass $m = 2\ \text{kg}$ rests on a smooth horizontal surface. A constant horizontal force $F = 10\ \text{N}$ directed along the displacement is applied to it. After travelling a distance $d = 5\ \text{m}$ the block has a speed $v$. Find $v$.

A body slides from rest down a smooth hill of height $h = 1.25\ \text{m}$ and enters a rough horizontal section with friction coefficient $\mu = 0.25$. What distance $s$ will the body travel before stopping? Take $g = 9.8\ \text{m/s}^2$.

A block of mass $m = 0.4\ \text{kg}$ is released from rest from a smooth hill of height $h = 0.8\ \text{m}$. It then crosses a rough horizontal section of length $L = 1.0\ \text{m}$ with friction coefficient $\mu = 0.2$ and compresses a massless spring of stiffness $k = 200\ \text{N/m}$. Find the maximum compression $x$ of the spring. Take $g = 9.8\ \text{m/s}^2$.

Self-check in a simulation (safe, no real moving bodies). Real ramps, jumps and fast carts are hazardous, so we work in a virtual lab such as PhET's "Energy Skate Park".

1. Switch to the no-friction mode, choose a U-shaped track and a starting point at height $h_1$.
2. Mark three points of the trajectory; at each one record the height $h$ and the speed $v$.
3. For each point compute $E_p = mgh$, $E_k = \frac{mv^2}{2}$ and the sum $E = E_k + E_p$. Check that $E$ is the same at all points, then plot $E_k$, $E_p$ and $E$ against height.
4. Turn friction on and repeat the measurements. Find the change in total mechanical energy $\Delta E$ and explain it by the sign and magnitude of the friction work $A_{\text{fric}}$.
5. Write your conclusion in three sentences: when $E$ is conserved, why it decreases with friction, and what role the initial height plays.

Show that in motion without friction the sum of kinetic and potential energy stays constant, while with friction the total mechanical energy decreases by exactly the work of the friction force.

A sketch of the track (a hill with a rise and a descent) with marked points 1, 2, 3 and the reference level $h = 0$; next to each point an energy bar chart: a bar for $E_k$, a bar for $E_p$ and a horizontal line at the level $E = E_k + E_p$. Below it, a graph of $E_k$, $E_p$ and $E$ versus the coordinate $x$ along the track: two curves that exchange height and a straight line for the constant sum; in the second graph (with friction) that straight line slopes gently downward, marking $\Delta E = A_{\text{fric}}$.

A car of mass $1200\ \text{kg}$ travels at $20\ \text{m/s}$ (about $72\ \text{km/h}$) and stops over $40\ \text{m}$ on a dry road. The initial kinetic energy is $K = \frac12 mv^2 = \frac12 \cdot 1200 \cdot 20^2 = 240\,000\ \text{J} = 240\ \text{kJ}$. Since the car stops, $\Delta K = -240\,000\ \text{J}$. The work done by the friction force over the stopping distance is $W = -F \cdot 40\ \text{m}$; setting $W = \Delta K$ gives $F = 6000\ \text{N}$.

This uses the work-energy theorem: the net work on the car equals the change in its kinetic energy, $W_{\text{net}} = \Delta K$. The friction force does negative work and removes the car's kinetic energy. No gravitational potential energy term appears because the road is level, so $U$ does not change.

A pole vaulter of mass $70\ \text{kg}$ sprints at $8\ \text{m/s}$ and plants the pole. Just before take-off the kinetic energy is $K = \frac12 \cdot 70 \cdot 8^2 = 2240\ \text{J}$. If all of this energy were converted to gravitational potential energy, the rise would be $h = \frac{K}{mg} = \frac{2240}{70 \cdot 9.8} \approx 3.3\ \text{m}$, so the vaulter's centre of mass could clear a bar around $3\ \text{m}$ above take-off height.

This uses conservation of mechanical energy: while the vaulter rises, $K + U$ stays constant, with $K$ shrinking and $U = mgh$ growing. The maximum rise follows from $K_{\text{before}} = mgh_{\text{max}}$. In a real vault the pole also stores elastic energy and the athlete does extra work, so the actual height differs slightly.

A phone of mass $0.20\ \text{kg}$ is dropped from a height of $1.5\ \text{m}$ in a durability test. At release the mechanical energy is all potential: $E = U = mgh = 0.20 \cdot 9.8 \cdot 1.5 = 2.94\ \text{J}$. Just before impact, if air resistance is negligible, all of it is kinetic: $K = \frac12 mv^2 = 2.94\ \text{J}$, giving $v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 1.5} \approx 5.4\ \text{m/s}$.

This uses conservation of mechanical energy when air resistance is negligible: the potential energy at the release height equals the kinetic energy just before impact, $mgh = \frac12 mv^2$. The sum $K + U$ stays constant at $2.94\ \text{J}$ throughout the fall.