Work, Energy and Power

A $2\text{kg}$ block slides at $6\text{m/s}$ onto a $4\text{kg}$ plank resting on frictionless ice. There is friction between block and plank. How much heat is generated by the time they move together?

The instinct is that you cannot answer without knowing $\mu$. You can — and $\mu$ never appears.

Momentum is conserved, since the ice exerts no horizontal force:

$v_{cm}=\frac{2(6)}{6}=2\text{m/s}$

$K_i=\frac{1}{2}(2)(36)=36\text{J},K_f=\frac{1}{2}(6)(4)=12\text{J}$

$Q=36-12=24\text{J}$

The coefficient of friction decides only how long it takes and how far each body slides — never how much heat appears. That is fixed the moment the initial velocities are.

The general result, which is worth knowing on sight:

$Q=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)v_{rel}^2=\frac{1}{2}\left(\frac{2\times 4}{6}\right)(36)=24\text{J}$

That bracket is the reduced mass, and this chapter is largely about the two ideas it encodes: heat depends on relative motion, and energy splits cleanly once you stand in the right frame.

1. Work done by a force is not work done on a system

At Main level $W=\vec{F}\cdot \vec{d}$ is unambiguous because bodies are rigid and forces act at one obvious place. Advanced problems break both assumptions.

$W=\int \vec{F}\cdot d\vec{r}\text{evaluated at the point of application}$

Three cases where the naive reading fails:

The rule that resolves all three: compute the work done by each force at its own point of application, and never assume two contacting surfaces have moved equally.

Illustration 1

A block is pulled $3\text{m}$ across a floor by a $10\text{N}$ horizontal force while friction of $4\text{N}$ opposes it. The floor is fixed. Find the work done by each force and the heat generated.

$W_{applied}=10(3)=+30\text{J}$

$W_{friction,onblock}=-4(3)=-12\text{J}$

$W_{friction,onfloor}=-4\times 0=0$, because the floor does not move.

$\Delta K=30-12=18\text{J}$

The heat is $4(3)=12\text{J}$ — and here it happens to equal the magnitude of the work done on the block, because the floor was stationary.

That coincidence is exactly what the next section destroys.

2. Friction heat depends on relative sliding

Two surfaces rub. The block advances $s_1$, the plank advances $s_2$, and the friction force has magnitude $f$ on each.

$W_{onblock}=-f{s}_1,W_{onplank}=+f{s}_2$

$Q=f{s}_1-f{s}_2=f\left(s_1-s_2\right)=f{s}_{rel}$

The heat is the friction force times the sliding distance at the interface, and the sliding distance is a relative displacement. Neither body's own journey is the right number.

This is the most reliably mis-answered idea in the chapter. A candidate who writes $Q=f{s}_1$ has computed the work done on the block, which is a different quantity, and the two agree only when the second surface is fixed.

Since the internal forces are a third-law pair, the total work they do is $-f{s}_{rel}$, which is never positive. Friction always destroys mechanical energy for the pair, even though it can do positive work on one member — as it does on the plank here.

Illustration 2

In the opening problem, take $\mu =0.5$ and $g=10$. Find the sliding distance at the interface and check the heat.

$f=\mu mg=0.5(2)(10)=10\text{N}$

$Q=f{s}_{rel}\Rightarrow 24=10s_{rel}\Rightarrow s_{rel}=2.4\text{m}$

Now halve the friction. With $\mu =0.25$, $f=5$ N and $s_{rel}=4.8$ m — the block slides twice as far along the plank and takes twice as long, but the heat is still 24 J. Momentum and energy fixed it before friction ever entered.

3. Force from a potential, in one and more dimensions

$F_x=-\frac{dU}{dx}⟶\vec{F}=-\nabla U=-\left(\frac{\partial U}{\partial x},\frac{\partial U}{\partial y},\frac{\partial U}{\partial z}\right)$

Advanced questions routinely hand you $U(x,y)$ and ask for the force, the equilibrium points, or their stability. The partial derivatives are the whole technique.

Equilibrium requires every component to vanish, not just one.

Stability in one dimension is read off the second derivative:

Illustration 3

A particle moves under $U(x,y)=3x^{2}y-y^3$ (SI units). Find the force at the point $(1,2)$.

$F_x=-\frac{\partial U}{\partial x}=-6xy$, and at $(1,2)$: $F_x=-12$ N

$F_y=-\frac{\partial U}{\partial y}=-\left(3x^2-3y^2\right)=3y^2-3x^2$, and at $(1,2)$: $F_y=12-3=+9$ N

$\vec{F}=(-12,9)\text{N},|\vec{F}|=15\text{N}$

The habit worth building: differentiate with respect to one variable while holding the other fixed, and take the minus sign once at the end rather than juggling it through the algebra.

4. Small oscillations about a minimum

Expand $U$ in a Taylor series about a minimum at $x_0$. The linear term vanishes because $U^{\prime }(x_0)=0$, so to leading order

$U(x)\approx U(x_0)+\frac{1}{2}{U}^{\prime \prime }(x_0){\left(x-x_0\right)}^2$

That is exactly a spring with $k_{eff}=U^{\prime \prime }(x_0)$. Therefore

$\omega =\sqrt{\frac{U^{\prime \prime }(x_0)}{m}}$

Every smooth potential minimum is a harmonic oscillator for small enough amplitude. This is why simple harmonic motion appears in so many unrelated systems, and it is a standard Advanced link between this chapter and Oscillations.

Illustration 4

A particle of mass $0.1\text{kg}$ moves under $U(x)=\frac{a}{x^2}-\frac{b}{x}$ with $a=2$, $b=4$ in SI units. Locate the equilibrium and find the frequency of small oscillations about it.

$\frac{dU}{dx}=-\frac{2a}{x^3}+\frac{b}{x^2}=0\Rightarrow x_0=\frac{2a}{b}=1\text{m}$

$\frac{d^{2}U}{d{x}^2}=\frac{6a}{x^4}-\frac{2b}{x^3}$, and at $x_0=1$: $12-8=4\text{N/m}$

Positive, so the equilibrium is stable, and

$\omega =\sqrt{\frac{4}{0.1}}=6.32\text{rad/s}\Rightarrow f=1.01\text{Hz}$

Recognise the potential: this is the standard model of a diatomic molecule's bond, and the calculation just done is how vibrational frequencies of molecules are actually estimated.

5. Energy in a non-inertial frame

A pseudo force is not a real force, but in the accelerating frame it does real work and must appear in the energy equation of that frame.

$W_{real}+W_{pseudo}=\Delta K_{frame}$

Both sides are frame-dependent. Kinetic energy is not invariant, work is not invariant, and only after choosing one frame and staying in it does the bookkeeping close.

Illustration 5

A lift accelerates upward at $a$. A block of mass $m$ is released from rest relative to the lift and falls a height $h$ as measured inside the lift. Find its speed relative to the lift on arrival, using the lift's frame.

Inside the lift, two forces do work: gravity $mg$ downward and the pseudo force $ma$ downward.

$W=\left(mg+ma\right)h=\frac{1}{2}m{v}_{rel}^2$

$v_{rel}=\sqrt{2(g+a)h}$

Read it as an effective gravity. An upward-accelerating lift behaves in every mechanical respect like a stronger gravitational field, $g_{eff}=g+a$. That single substitution handles pendulums, projectiles and falling bodies inside accelerating vehicles.

Check the extreme: in free fall, $a=-g$, so $g_{eff}=0$ and the block never moves relative to the lift. Correct.

6. Springs with both ends free

When both ends of a spring move, the stored energy depends on the change in length, which is a relative displacement again:

$U_{spring}=\frac{1}{2}k{\left(\Delta \ell \right)}^2$

The key structural fact for two-block problems: the spring is at maximum compression or extension exactly when the two blocks have the same velocity, because that is the instant the separation stops changing.

At that instant, momentum conservation gives the common velocity immediately, and energy conservation then gives the deformation.

Illustration 6

Blocks of $2\text{kg}$ and $3\text{kg}$ on a frictionless floor are joined by a spring of stiffness $200\text{N/m}$, initially at natural length. The $2\text{kg}$ block is given $5\text{m/s}$ toward the other. Find the maximum compression.

Common velocity at maximum compression:

$v=\frac{2(5)}{5}=2\text{m/s}$

Energy:

$\frac{1}{2}(2)(25)=\frac{1}{2}(5)(4)+\frac{1}{2}(200)x^2$

$25=10+100x^2\Rightarrow x^2=0.15\Rightarrow x=0.387\text{m}$

The shortcut worth learning. The energy that goes into the spring is exactly the internal kinetic energy, $\frac{1}{2}\mu v_{rel}^2$ with $\mu =\frac{2(3)}{5}=1.2$:

$\frac{1}{2}(1.2)(25)=15\text{J}=\frac{1}{2}(200)x^2\Rightarrow x=0.387\text{m}$

Same answer in one line, and it makes clear why the maximum compression does not depend on the frame.

7. The centre-of-mass frame, and König's theorem

Split any system's kinetic energy into the motion of the centre of mass and the motion about it:

$K=\underset{\text{translational}}{\underbrace{\frac{1}{2}M{v}_{cm}^2}}+\underset{\text{about the c.m.}}{\underbrace{K_{internal}}}$

For two bodies the internal part takes a remarkably clean form:

$K_{internal}=\frac{1}{2}\mu v_{rel}^2,\mu =\frac{m_1{m}_2}{m_1+m_2}$

The first term is untouchable. With no external force, $v_{cm}$ is constant, so $\frac{1}{2}M{v}_{cm}^2$ can never be converted into anything. Only the internal part is available to become heat, spring energy or deformation.

That single observation explains the opening problem, the spring problem, and every collision at once. It is the most powerful idea in this chapter.

Illustration 7

Two particles of $3\text{kg}$ and $6\text{kg}$ move at $4\text{m/s}$ and $1\text{m/s}$ in the same direction. Find the total kinetic energy, and split it into translational and internal parts.

$K_{total}=\frac{1}{2}(3)(16)+\frac{1}{2}(6)(1)=24+3=27\text{J}$

$v_{cm}=\frac{3(4)+6(1)}{9}=\frac{18}{9}=2\text{m/s}$

$K_{trans}=\frac{1}{2}(9)(4)=18\text{J}$

$\mu =\frac{3(6)}{9}=2$, and $v_{rel}=3\text{m/s}$, so $K_{int}=\frac{1}{2}(2)(9)=9\text{J}$

$18+9=27\text{J}$. The split checks out.

At most $9\text{J}$ can ever be dissipated, however violent the interaction. If they stick together, exactly that $9$ J becomes heat and $18$ J survives untouched.

8. Collisions, seen from the centre of mass

In the CM frame the total momentum is zero, so the two momenta are always equal and opposite. That constraint makes the whole analysis nearly trivial:

The general energy loss follows immediately. Only the internal kinetic energy is available, and a coefficient of restitution $e$ leaves a fraction $e^2$ of it:

$\Delta K=\frac{1}{2}\mu v_{rel}^2\left(1-e^2\right)$

Setting $e=1$ gives zero loss and $e=0$ gives the full $\frac{1}{2}\mu v_{rel}^2$ — both correct, and both recovered from one formula.

For an oblique collision between smooth bodies, apply $e$ only along the line of impact. The tangential components pass through untouched, because no impulse acts along the common tangent.

Illustration 8

A $2\text{kg}$ ball at $10\text{m/s}$ strikes a stationary $3\text{kg}$ ball head-on with $e=0.6$. Find the energy lost.

$\mu =\frac{2(3)}{5}=1.2$, $v_{rel}=10\text{m/s}$

$\Delta K=\frac{1}{2}(1.2)(100)\left(1-0.36\right)=60(0.64)=38.4\text{J}$

Check against the direct route. The final velocities are $v_1=0.4$ and $v_2=6.4$ m/s, so

$K_f=\frac{1}{2}(2)(0.16)+\frac{1}{2}(3)(40.96)=0.16+61.44=61.6\text{J}$

$K_i=100$ J, so the loss is $38.4$ J. The formula and the arithmetic agree, and the formula took one line.

Illustration 9

A ball strikes a smooth floor at $30°$ to the horizontal with speed $u$ and $e=0.5$. Find its rebound speed and angle.

The line of impact is vertical, so restitution applies only to the vertical component.

$v=\sqrt{(0.866u{)}^2+(0.25u{)}^2}=0.901u$

$\tan {\theta }^{\prime }=\frac{0.25}{0.866}\Rightarrow {\theta }^{\prime }=16.1°$

The ball comes off flatter than it went in — $16°$ against $30°$ — because only the vertical component was reduced. That flattening is why a bouncing ball's trajectory degenerates into a skid.

9. Power, and where it goes

$P=\vec{F}\cdot \vec{v}\text{at the point of application}$

For internal friction between two bodies, the rate of heat production follows the same relative rule as the total:

$\frac{dQ}{dt}=f{v}_{rel}$

Illustration 10

In the opening block-and-plank problem, find the rate of heat production at the instant the block moves at $5\text{m/s}$.

Momentum: $2(6)=2(5)+4v_2\Rightarrow v_2=0.5\text{m/s}$

$v_{rel}=4.5\text{m/s}$, and with $\mu =0.5$ the friction is $10$ N:

$\frac{dQ}{dt}=10(4.5)=45\text{W}$

Read the trend: as the two speeds converge, $v_{rel}\to 0$ and the heating rate falls to zero, even though the friction force is constant throughout. All 24 J is delivered, but at an ever-decreasing rate — which is why the final approach to common velocity takes disproportionately long.

Summary

• Friction heat is $f{s}_{rel}$ — the force times the sliding at the interface, never times either body's own displacement.
• Total heat is fixed by momentum and energy alone; $\mu$ decides only how far and how long, not how much.
• $W=\int \vec{F}\cdot d\vec{r}$ is evaluated at the point of application.
• $\vec{F}=-\nabla U$. Equilibrium needs every component of the gradient to vanish.
• Every smooth minimum is a spring: $k_{eff}=U^{\prime \prime }(x_0)$ and $\omega =\sqrt{U^{\prime \prime }/m}$.
• Pseudo forces do real work in their own frame. An upward-accelerating lift is just $g_{eff}=g+a$.
• A spring between two blocks is at maximum deformation exactly when their velocities are equal.
• König: $K=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}\mu v_{rel}^2$. Only the second term can ever be dissipated.
• In the CM frame, elastic collisions preserve both speeds and merely rotate them; perfectly inelastic ones bring both to rest.
• $\Delta K=\frac{1}{2}\mu v_{rel}^2\left(1-e^2\right)$ covers every collision from elastic to perfectly inelastic.
• Oblique impact: apply $e$ only along the line of impact; tangential components survive untouched.
• $dQ/dt=f{v}_{rel}$, so heating stops when the relative sliding does.