Conservation of energy; conservative and non-conservative forces

Conservative vs non-conservative. A force is conservative if the work it does moving a body between two points is independent of the path taken — equivalently, the work done around any closed loop is zero. Gravity and the spring force are conservative; friction and air resistance are non-conservative, because the work done against them depends on the actual path length (a longer, winding path loses more energy to friction, even between the same endpoints).

Potential energy from a conservative force. For every conservative force we can define an associated potential energy $U$, such that the work it does equals the decrease in $U$:

where: $W_{conservative}$ is the work done by the conservative force, and $U_i$, $U_f$ are the potential energies at the initial and final positions. For gravity near Earth's surface, $U=mgh$ (with $h$ measured from a chosen reference level); for a spring stretched or compressed by $x$, $U=\tfrac12 kx^2$.

Conservation of mechanical energy. When only conservative forces do work (no friction, no external drag), total mechanical energy — kinetic plus potential — stays fixed:

where: $K=\tfrac12 mv^2$ is kinetic energy and $U$ is potential energy. This follows from the work–energy theorem $W_{net}=\Delta K$: if $W_{net}$ is entirely due to conservative forces, $W_{net}=-\Delta U$, so $\Delta K=-\Delta U$, i.e. $\Delta(K+U)=0$. The figure below is the standard way this is tested: plot $K$, $U$, and $E$ against height during a fall. $E$ is a flat horizontal line; $K$ and $U$ trade off exactly, so wherever $U$ is maximum $K=0$ (and $E=U_{max}$), and wherever $U=0$, $K$ is maximum (and $E=K_{max}$).

When non-conservative forces like friction also act, mechanical energy is not conserved — some converts to heat — and although $W_{net}=\Delta K$ still holds in general, $\Delta K\neq -\Delta U$ once friction is present.

A block of mass $2\ \text{kg}$ slides down a smooth (frictionless) curved track from a height of $5\ \text{m}$ and then moves across a rough horizontal section, where a constant friction force of $4\ \text{N}$ acts. Find (a) the block's speed at the bottom of the curve, (b) the distance it travels on the rough section before stopping, and (c) the coefficient of kinetic friction on that section.

Step 1. On the smooth curve, only gravity does work, so mechanical energy is conserved. Taking the bottom as the reference level, $E=mgh=(2)(9.8)(5)=98\ \text{J}$, all of it kinetic at the bottom: $\tfrac12 mv^2=98 \implies v=\sqrt{98}\approx 9.9\ \text{m/s}$.

Step 2. On the rough section, friction removes kinetic energy until the block stops. By the work–energy theorem, $-fS=0-98$, so $S=\dfrac{98}{f}=\dfrac{98}{4}=24.5\ \text{m}$.

Step 3. For the coefficient of kinetic friction, use $f=\mu mg$ on the level section, where the normal force equals $mg$: $\mu=\dfrac{f}{mg}=\dfrac{4}{(2)(9.8)}\approx 0.204$.

The block covers $24.5\ \text{m}$ before stopping, with a coefficient of kinetic friction of about $0.204$; all $98\ \text{J}$ of mechanical energy is accounted for as heat generated by friction — no energy is truly lost, only converted out of the K+U form.

• Believing energy is "lost" when friction acts — mechanical energy decreases, but total energy including heat is always conserved; only mechanical energy needs purely conservative forces to stay constant.
• Assuming every force is conservative — normal reaction and tension often do zero work (perpendicular to motion), but friction genuinely removes mechanical energy whenever it acts.
• Applying $E=K+U=\text{constant}$ on a section where friction or an applied force does work, without first checking that only conservative forces act there.
• Forgetting to fix a single, consistent reference level for $U$ before computing energies — only differences in $U$ matter, but the same reference must be used throughout one problem.

Ask the tutor: Want to try a few more conservation-of-energy problems on inclines, springs, or pendulums to build speed for the exam?