Conservation of Energy

A shortcut past the middle of the problem

A child starts at rest at the top of a slide 3.2 m high and arrives at the bottom moving. If you attack that with forces you need the shape of the slide, the angle at every point, the normal force at every point, and an integration you cannot do by hand. Yet the answer is one line long, and it does not depend on the shape of the slide at all.

That is what conservation of energy buys you. It ignores the middle of the problem. You take a snapshot at the start, a snapshot at the end, and demand that one number match. Everything that happened in between (the curving, the speeding up, the changing direction) never has to be described.

The previous lesson, conservative forces and potential energy, built the tool: a force is conservative when the work it does depends only on where you start and where you finish, and for such a force you can define a potential energy $U$ whose drop equals the work done. Gravity near the ground gives $U_g = mgy$. An ideal spring gives $U_s = \tfrac12 k x^2$. This lesson is about using those to run a problem from end to end.

The ledger

Start from the work-energy theorem, which you already have: the total work done on an object equals the change in its kinetic energy,

$W_{\text{total}} = \Delta K = \tfrac12 m v_f^2 - \tfrac12 m v_i^2 .$

Here $K = \tfrac12 m v^2$ is the kinetic energy in joules, $m$ the mass in kg, $v$ the speed in m/s. Now split the total work into two piles: work done by conservative forces, and work done by everything else.

$W_{\text{cons}} + W_{\text{other}} = \Delta K .$

The whole point of a potential energy is that $W_{\text{cons}} = -\Delta U$. Gravity doing $+50$ J of work on a falling rock is the same statement as the gravitational potential energy dropping by 50 J. Substitute that in and move it across:

$W_{\text{other}} = \Delta K + \Delta U .$

Define the mechanical energy $E = K + U$, the sum of kinetic and all potential energies. Then

$\boxed{\,E_f = E_i + W_{\text{other}}\,}$

Read that as an account statement. $E_i$ is what you start with. $W_{\text{other}}$ is everything deposited or withdrawn by forces that have no potential energy: friction, air drag, a hand pushing, a rope being pulled by a motor. $E_f$ is what you end with. When no such force does work, $W_{\text{other}} = 0$ and

$\tfrac12 m v_i^2 + U_i = \tfrac12 m v_f^2 + U_f .$

Two snapshots, one equation. Notice what is not in it: time, distance along the path, the shape of the track, or the direction of motion. Energy is a scalar. It has no components and no signs to chase.

Worked example 1. A 25 kg child starts from rest at the top of a frictionless slide 3.2 m high. How fast is she moving at the bottom?

Put $y = 0$ at the bottom. Snapshot one: $v_i = 0$, $y_i = 3.2$ m. Snapshot two: $v_f = ?$, $y_f = 0$. Nothing but gravity and the normal force acts, and the normal force is perpendicular to the motion so it does no work. So

$\tfrac12 m v_i^2 + m g y_i = \tfrac12 m v_f^2 + m g y_f$

$0 + m g (3.2) = \tfrac12 m v_f^2 + 0 .$

Every term has an $m$, so cancel it. This happens constantly, and it is why the mass of the child was never needed.

$v_f = \sqrt{2 g h} = \sqrt{2(9.8)(3.2)} = \sqrt{62.72} = 7.92\ \text{m/s}.$

The same single line handles a dropped stone ($h = 5.0$ m gives 9.90 m/s), a pendulum bob released 0.45 m above its lowest point (2.97 m/s), and a skateboarder who enters a bowl at 8.0 m/s and rises until all her kinetic energy is potential: $h = v^2/2g = 64/19.6 = 3.27$ m. Different pictures, one equation.

When energy leaks out

$W_{\text{other}}$ is usually friction, and friction is the one non-conservative force you will meet constantly. A kinetic friction force of magnitude $f_k = \mu_k N$ opposes the motion at every instant, so over a path of length $d$ it does work $-f_k d$. The minus sign is the whole story: friction always withdraws from the account.

Notice the $d$. Unlike gravity, friction cares about the length of the path, not the displacement. Take a longer route and you lose more. That is exactly the property that makes it non-conservative, and it means you cannot define a "friction potential energy".

Worked example 2. A 1.0 kg block slides down a frictionless ramp from 2.0 m high, then crosses a rough floor with $\mu_k = 0.40$. How far does it go before stopping?

Take the two snapshots as far apart as possible: at the top of the ramp, and where the block stops. Both have $v = 0$, so both kinetic energies are zero. There is no need to find the speed at the bottom of the ramp at all.

$E_f - E_i = W_{\text{other}}$ $(0 + 0) - (0 + mgh) = -\mu_k m g d$ $mgh = \mu_k m g d \quad\Rightarrow\quad d = \frac{h}{\mu_k} = \frac{2.0}{0.40} = 5.0\ \text{m}.$

On the flat floor the normal force is $N = mg$, which is why $f_k = \mu_k mg$. On an incline it is not: there $N = mg\cos\theta$, and forgetting the cosine is the most common way this goes wrong.

Worked example 3. A 2.0 kg block is released from rest on a $35^\circ$ incline, slides 1.5 m down against $\mu_k = 0.20$, and then compresses a spring of stiffness $k = 2000$ N/m mounted along the incline. Find the maximum compression $x$.

Take snapshot one at release and snapshot two at maximum compression. At maximum compression the block is momentarily at rest, so both kinetic energies are zero again. Between them the block travels a total distance $1.5 + x$ along the incline, dropping a height $(1.5 + x)\sin 35^\circ$, and friction acts over that whole distance including the part inside the spring. That "$+x$" is the part people drop.

Gravity releases $mg\sin\theta\,(1.5+x)$. With $mg = 19.6$ N and $\sin 35^\circ = 0.5736$, that coefficient is $19.6 \times 0.5736 = 11.24$ N. Friction takes $\mu_k m g\cos\theta\,(1.5+x)$, and with $\cos 35^\circ = 0.8192$ that coefficient is $0.20 \times 19.6 \times 0.8192 = 3.21$ N. The spring stores $\tfrac12 k x^2$. So

$(11.24 - 3.21)(1.5 + x) = \tfrac12 (2000) x^2$ $8.03(1.5 + x) = 1000 x^2$ $1000x^2 - 8.03x - 12.05 = 0 .$

Quadratic formula: $x = \dfrac{8.03 + \sqrt{8.03^2 + 4(1000)(12.05)}}{2000} = \dfrac{8.03 + \sqrt{64.5 + 48200}}{2000} = \dfrac{8.03 + 219.6}{2000} = 0.114\ \text{m}.$

Take the positive root; the negative one describes a spring stretched the wrong way, which the setup does not allow.

When the track pushes back: loops and spheres

Some problems need one more equation, because energy alone cannot tell you whether an object is still touching the track. Energy gives you speed; contact is a statement about the normal force, and for that you need Newton's second law in the radial direction.

At the top of a vertical loop of radius $R$, both gravity and the normal force point down, toward the centre, and together they supply the centripetal force:

$N + mg = \frac{m v_{\text{top}}^2}{R}.$

The track can only push, never pull, so $N \ge 0$. The slowest the object can go and still be held on the circle is the case $N = 0$, where gravity alone does the job:

$v_{\text{top}}^2 = gR .$

That is the condition. Now, and only now, bring in energy to find where it must start. Releasing from rest at height $h$ above the bottom of the loop, on a frictionless track:

$mgh = \tfrac12 m v_{\text{top}}^2 + mg(2R) .$

Substitute $v_{\text{top}}^2 = gR$, cancel $m$, and divide by $g$:

$h = \tfrac12 R + 2R = 2.5R .$

Worked example 4. For a loop of radius 2 m with $g = 10$ m/s², the minimum release height is $h = 2.5(2) = 5$ m. For $R = 2.5$ m it is 6.25 m. The answer is always $2.5R$ and never involves the mass.

If friction dissipates a known amount $Q$ around the loop, the ledger just gains one term. A 0.40 kg bead needing to clear a loop of $R = 0.50$ m while losing 0.30 J to friction needs, at the bottom,

$\tfrac12 m v_b^2 = \tfrac12 m (gR) + mg(2R) + Q = \tfrac12(0.40)(9.8)(0.50) + (0.40)(9.8)(1.0) + 0.30$ $\tfrac12(0.40)v_b^2 = 0.98 + 3.92 + 0.30 = 5.20 \quad\Rightarrow\quad v_b^2 = 26.0,\ v_b = 5.10\ \text{m/s}.$

The block sliding off a sphere is the same idea turned inside out. On the outside of a sphere of radius $R$, gravity points partly inward and the normal force points outward, so at an angle $\theta$ from the top,

$mg\cos\theta - N = \frac{mv^2}{R},$

and the block leaves when $N = 0$, that is when $v^2 = gR\cos\theta$. Energy supplies the other relation: it has fallen a height $R(1 - \cos\theta)$, so $v^2 = 2gR(1-\cos\theta)$. Set them equal:

$2gR(1 - \cos\theta) = gR\cos\theta \quad\Rightarrow\quad \cos\theta = \tfrac23 .$

The block always leaves at the height $\tfrac23 R$ above the centre, whatever $R$ and whatever $m$. For $R = 1.6$ m that is 1.07 m. Its speed there is $v = \sqrt{gR\cos\theta} = \sqrt{2gR/3}$; for $R = 2.4$ m, $v = \sqrt{2(9.8)(2.4)/3} = \sqrt{15.68} = 3.96$ m/s.

More than one moving thing

When several objects move, put every one of them in the same ledger. The kinetic energy term becomes a sum, and so does the potential energy term.

Worked example 5. A 3.0 kg block on a frictionless table is joined over a massless pulley to a 2.0 kg hanging block. Find their common speed after the hanging block descends 0.80 m from rest.

The rope is inextensible, so both blocks always move at the same speed $v$. Only the hanging block changes height. The rope tension does positive work on one block and exactly the same amount of negative work on the other, so it cancels out of the total, which is why treating the two blocks as one system is worth doing.

$m_2 g d = \tfrac12 (m_1 + m_2) v^2$ $(2.0)(9.8)(0.80) = \tfrac12 (5.0) v^2$ $15.68 = 2.5 v^2 \quad\Rightarrow\quad v^2 = 6.272,\ v = 2.50\ \text{m/s}.$

Two more wrinkles show up in harder versions of this. First, if the system can recoil (a block sliding down a wedge that is itself free to slide) momentum in the free direction is conserved too, and you need both equations, because the wedge carries away some of the kinetic energy. Second, for an extended object like a chain or a rope, the gravitational potential energy is $Mgy_{\text{cm}}$, using the height of the centre of mass.

Worked example 6. A uniform chain of length 1.8 m lies along a frictionless table with one end at the edge, and slides off over a smooth peg. How fast is it moving when the last link leaves the table?

You cannot track the links one at a time. But you do not have to: the whole chain always moves at one speed, so $K = \tfrac12 M v^2$ no matter how it is draped. Initially it lies flat on the table, so its centre of mass is at table height, $y_{\text{cm},i} = 0$. Finally it hangs straight down from the edge, a length 1.8 m, so its centre of mass is at $-0.9$ m. The drop is $L/2$:

$Mg\frac{L}{2} = \tfrac12 M v^2 \quad\Rightarrow\quad v = \sqrt{gL} = \sqrt{(9.8)(1.8)} = \sqrt{17.64} = 4.2\ \text{m/s}.$

The mass cancels again, and the answer is just the free-fall speed for a drop of $L/2$.

How to recognise that this is the technique

Reach for energy conservation when a problem gives you two states and asks for a speed, a height, a compression, or a distance connecting them, and says nothing about time. The giveaway phrases: "released from rest", "how fast at the bottom", "how high does it rise", "maximum compression", "how far before it stops". If the word time or acceleration appears in the question, energy is probably the wrong tool and kinematics or $F = ma$ is the right one.

A second signal is a track whose shape you are not told. A curved slide, a bowl, a ski jump, a rope swing: the problem is deliberately withholding the geometry because the answer does not need it. That is energy telling you what it is.

The procedure, every time: pick your two snapshots and put them as far apart as you can, choose and write down $y = 0$, list every energy term at each snapshot including the zeros, then add $W_{\text{other}}$ for friction or an applied force. If the problem asks whether something stays on a track, you will need one extra line from Newton's second law in the radial direction. If it contains a collision or a sudden grab, split the problem at that instant and use momentum across it.

What energy will never tell you is how long anything took, or how hard you had to work per second to do it. That question (the rate of energy transfer) is what power is for.