Unit 3: Work, Energy, and Power

Useful variables

• $W$ = work (Units: joules, $\text{J}$)
• $K$ = kinetic energy (Units: $\text{J}$)
• $U$ = potential energy (Units: $\text{J}$)
• $E$ = total mechanical energy (Units: $\text{J}$)
• $P$ = power (Units: watts, $\text{W} = \text{J/s}$)
• $\vec{F}$ = force
• $\vec{r}$ = position vector
• $x$ = one-dimensional position or displacement
• $k$ = spring constant

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Work

Work measures energy transferred by a force acting through a displacement. For a constant force,

$$W = \vec{F}\cdot \Delta \vec{r} = F\Delta r\cos\theta.$$

Only the component of force parallel to displacement does work. A perpendicular force can change direction without changing speed, so it does no work at that instant. Due to this property, perpendicular forces like the normal force cannot exert any work.

For a variable force, use the line integral (just the integral over the path of an object) to evaluate work:

$$W = \int_C \vec{F}\cdot d\vec{r}.$$

In one dimension (which involves most AP Physics C problems) this becomes

$$W = \int_{x_i}^{x_f} F_x(x)\, dx,$$

the signed area under the force-position graph.

The geometry of the dot product

Work is a scalar, so it has a sign but no direction. Since $W = \vec{F}\cdot\Delta\vec{r} = F\,\Delta r\cos\theta$, the sign of the work is set entirely by the angle $\theta$ between the force and the displacement:

• Positive work ($0\le\theta<90^\circ$): the force has a component along the motion and speeds the object up (it transfers energy into the object). A horizontal push on a sliding box does positive work.
• Negative work ($90^\circ<\theta\le180^\circ$): the force opposes the motion and slows the object (it removes energy). Kinetic friction on a sliding box does negative work.
• Zero work ($\theta=90^\circ$): a force perpendicular to the velocity does no work. The normal force on a block sliding along a floor, the tension on a ball in uniform circular motion, and the magnetic force on a charge all do zero work even though they are nonzero forces.

Tip

Tip

Positive work will always speeds up the object, negative work always slows down the object, and zero work causes the speed to remain constant, even though the direction might change.

Example. A force directed along the $x$-axis varies with position as follows: it is constant at $F=20\ \text{N}$ from $x=0$ to $x=3\ \text{m}$, then ramps linearly down to $0\ \text{N}$ at $x=5\ \text{m}$. Find the work done from $x=0$ to $x=5\ \text{m}$.

The work is the area under the graph. Split it into a rectangle and a triangle:

$$W = \underbrace{(20\ \text{N})(3\ \text{m})}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}(2\ \text{m})(20\ \text{N})}_{\text{triangle}} = 60\ \text{J} + 20\ \text{J} = 80\ \text{J}.$$

If the force had pointed in the $-x$ direction over some interval, that area would count as negative. The graph method and the integral $\int F_x\,dx$ are the same calculation; the graph just makes the geometry visible.

Example. A position-dependent force $F(x)=\alpha x^2-\beta x$ acts along the $x$-axis, where $\alpha=6.0\ \text{N/m}^2$ and $\beta=10\ \text{N/m}$. Find the work it does on a particle moving from $x_i=0.50\ \text{m}$ to $x_f=3.0\ \text{m}$, and state whether the force adds or removes mechanical energy overall.

Use $W = \int_{x_i}^{x_f} F(x)\,dx$:

$$W=\int_{0.50}^{3.0}(\alpha x^2-\beta x)\,dx =\left[\frac{\alpha x^3}{3}-\frac{\beta x^2}{2}\right]_{0.50}^{3.0}.$$

Substitute values:

$$W=\left[2x^3-5x^2\right]_{0.50}^{3.0} =(54-45)-\left(0.25-1.25\right)=10\ \text{J}.$$

Even though the force is negative over part of the interval and positive later, the net work is positive, so it adds $10\ \text{J}$ of kinetic energy. Whenever the force is not constant, you cannot use $W = Fd$; you must integrate.

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Kinetic Energy and the Work-Energy Theorem

Kinetic energy is defined as the energy of motion. Translational kinetic energy of a particle is defined as

$$K = \frac{1}{2}mv^2.$$

An important relationship between kinetic energy and work is the Work-Energy Theorem.

Theorem (Work-Energy Theorem). The net work done on a particle equals the change in its kinetic energy, $W_{\text{net}} = \Delta K$.

Proof (Work-Energy Theorem). Start with Newton’s second law and the definition of work:

$$W_{\text{net}} = \int \vec{F}_{\text{net}}\cdot d\vec{r} = \int m\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = \int m\vec{v}\cdot d\vec{v} = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2.$$

Therefore,

$$W_{\text{net}}=\Delta K.$$

Example. A block of mass $m = 2.0\ \text{kg}$ slides across a level floor with initial speed $v_0 = 6.0\ \text{m/s}$. The coefficient of kinetic friction is $\mu_k = 0.30$. How far does it slide before stopping? Use $g = 9.8\ \text{m/s}^2$.

The only horizontal force is kinetic friction, $f_k = \mu_k F_N = \mu_k mg$, directed opposite the motion. Over a distance $d$ it does negative work

$$W_{\text{net}} = -f_k d = -\mu_k mg\, d.$$

By the work-energy theorem, $W_{\text{net}} = \Delta K = 0 - \tfrac{1}{2}mv_0^2$:

$$-\mu_k mg\, d = -\tfrac{1}{2}mv_0^2.$$

The mass cancels, so the stopping distance does not depend on $m$:

$$d = \frac{v_0^2}{2\mu_k g} = \frac{(6.0)^2}{2(0.30)(9.8)} = \frac{36}{5.88} \approx 6.1\ \text{m}.$$

Notice that the stopping distance scales with $v_0^2$: doubling the speed quadruples the distance.

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Conservative Forces and Potential Energy

Definition. A force is conservative if its work depends only on the initial and final positions, not on the path taken. Equivalently,

$$\oint \vec{F}\cdot d\vec{r} = 0$$

around any closed path. The integral (known as a surface integral) represents the integral around a path, and thus represents the fact that in a closed loop the work done is $0$.

For a conservative force, define potential energy $U$ by

$$W_{\text{cons}} = -\Delta U.$$

In one dimension,

$$F_x = -\frac{dU}{dx}.$$

This comes directly from comparing a tiny amount of conservative work to a tiny change in potential energy:

$$dW=F_x\,dx,$$

and

$$dW=-dU.$$

Therefore

$$F_x\,dx=-dU,$$

so

$$F_x=-\frac{dU}{dx}.$$

In three dimensions,

$$\vec{F} = -\nabla U.$$

The $\nabla$ symbol is just an extension of a derivative to all three dimensions. Potential energy is not an absolute property; it requires a reference level, which is usually set at some point at infinity or zero. Only changes in potential energy affect mechanics.

Gravitational Potential Energy

Near Earth’s surface, where $g$ is approximately constant, the gravitational potential energy of an object is

$$U_g = mgy$$

if $U_g = 0$ is chosen at $y=0$. The change in gravitational potential energy is

$$\Delta U_g = mg\Delta y.$$

For universal gravitation, the usual zero point is defined at infinity, resulting in:

$$U_g(r) = -\frac{GMm}{r}.$$

The negative sign means a bound mass has less energy than it would have infinitely far away.

Proof (Near-Earth and Universal Gravitational Potential Energy). Near Earth’s surface, the gravitational force is approximately constant:

$$\vec{F}_g=-mg\hat{y}.$$

Since $W_g=-\Delta U_g$,

$$\Delta U_g=-W_g=-\int_{y_i}^{y_f}(-mg)\,dy=mg(y_f-y_i).$$

So, choosing $U_g=0$ at $y=0$,

$$U_g=mgy.$$

For universal gravitation,

$$F_r=-\frac{GMm}{r^2}.$$

Using $F_r=-dU/dr$,

$$\frac{dU}{dr}=\frac{GMm}{r^2}.$$

Integrate:

$$U(r)=-\frac{GMm}{r}+C.$$

Choosing $U(\infty)=0$ forces $C=0$, so

$$U_g(r)=-\frac{GMm}{r}.$$

The fact that we could even define a potential energy depends on gravity being conservative: the work it does between two points does not depend on the route taken.

Proof (gravity near Earth is path-independent). Near Earth’s surface $\vec{F}_g = -mg\,\hat{y}$, a constant vector. For any path from point $A$ to point $B$,

$$W_g=\int_A^B \vec{F}_g\cdot d\vec{r}.$$

Since $\vec{F}_g=-mg\hat{y}$ and $d\vec{r}=dx\hat{x}+dy\hat{y}$,

$$W_g=\int_A^B (-mg\hat{y})\cdot(dx\hat{x}+dy\hat{y}).$$

The $x$ part vanishes because $\hat{y}\cdot\hat{x}=0$, leaving

$$W_g=-mg\int_{y_A}^{y_B}dy=-mg(y_B-y_A).$$

The $x$-displacement drops out because $\hat{y}\cdot\hat{x}=0$, so only the change in height matters. A box carried straight up, or up a long ramp, or along a wiggling staircase to the same final height, all involve the same gravitational work. Around any closed loop ($y_B = y_A$), $W_g = 0$, which is the defining property of a conservative force.

Example. With what speed $v_{\text{esc}}$ must a projectile leave a planet’s surface (mass $M$, radius $R$, no air) so that it just barely reaches infinity (aka escapes the gravitational pull of the planet)? Use $U_g(r) = -GMm/r$.

“Just barely reaches infinity” means the projectile arrives at $r\to\infty$ with zero speed. With only gravity acting, mechanical energy is conserved:

$$\tfrac{1}{2}mv_{\text{esc}}^2 + \left(-\frac{GMm}{R}\right) = \underbrace{0}_{K_\infty} + \underbrace{0}_{U_\infty}.$$

Solving for $v_{\text{esc}}$,

$$\tfrac{1}{2}mv_{\text{esc}}^2 = \frac{GMm}{R}, \qquad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}.$$

The mass of the projectile cancels, so escape speed is the same for a pebble or a rocket. Using $g = GM/R^2$ at the surface, this can be rewritten as $v_{\text{esc}} = \sqrt{2gR}$. For Earth ($g = 9.8\ \text{m/s}^2$, $R = 6.37\times10^6\ \text{m}$), $v_{\text{esc}}\approx 1.12\times10^4\ \text{m/s}$, about $11.2\ \text{km/s}$.

Spring Potential Energy

For an ideal, massless spring, the potential energy stored in the spring is

$$U_s = \frac{1}{2}kx^2,$$

where $x$ is displacement from equilibrium.

Proof (Spring Potential Energy). Hooke’s law is

$$F_s=-kx.$$

For a conservative force,

$$F_s=-\frac{dU_s}{dx}.$$

Therefore

$$-kx=-\frac{dU_s}{dx},$$

so

$$\frac{dU_s}{dx}=kx.$$

Integrating gives

$$U_s=\frac{1}{2}kx^2+C.$$

Choosing $U_s=0$ at equilibrium, where $x=0$, makes $C=0$. Thus

$$U_s=\frac{1}{2}kx^2.$$

Example. A spring has stiffness $k = 400\ \text{N/m}$. How much work must an external agent do to stretch it from its natural length to $x_1 = 0.10\ \text{m}$, and then how much additional work to stretch it from $x_1$ to $x_2 = 0.20\ \text{m}$?

To stretch the spring slowly, the external force must balance the spring force, so $F_{\text{ext}}(x) = +kx$. The work done by this external force is

$$W_{\text{ext}} = \int_{0}^{x} kx'\,dx' = \tfrac{1}{2}kx^2,$$

which is exactly the stored potential energy. For the first stretch,

$$W_1 = \tfrac{1}{2}(400)(0.10)^2 = 2.0\ \text{J}.$$

To reach $x_2 = 0.20\ \text{m}$, the total stored energy is

$$\tfrac{1}{2}(400)(0.20)^2 = 8.0\ \text{J},$$

so the additional work is

$$W_2 = 8.0\ \text{J} - 2.0\ \text{J} = 6.0\ \text{J}.$$

Stretching the second $10\ \text{cm}$ takes three times the work of the first, because the force grows with displacement — the energy goes as $x^2$, not $x$. The spring itself does work $-\tfrac{1}{2}kx^2$ during stretching (opposing the motion).

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Conservation of Mechanical Energy

Defined mechanical energy as

$$E_{\text{mech}} = K + U.$$

If only conservative forces do work (e.g. no friction), mechanical energy is conserved:

$$K_i + U_i = K_f + U_f.$$

Proof (Conservation of Mechanical Energy). The work-energy theorem says

$$W_{\text{net}}=\Delta K.$$

If only conservative forces do work,

$$W_{\text{net}}=W_{\text{cons}}.$$

By definition of potential energy,

$$W_{\text{cons}}=-\Delta U.$$

Therefore

$$\Delta K=-\Delta U,$$

or

$$\Delta(K+U)=0.$$

So

$$K_i+U_i=K_f+U_f.$$

Example. A bob on a string of length $L = 1.5\ \text{m}$ is released from rest at an angle $\theta = 60^\circ$ from vertical. Find its speed at the lowest point. Ignore air resistance.

The tension is always perpendicular to the bob’s velocity, so it does no work; only gravity does work, and mechanical energy is conserved. The bob’s height above the lowest point when the string makes angle $\theta$ is

$$h = L - L\cos\theta = L(1-\cos\theta).$$

Taking the lowest point as $U = 0$ and using $K_i = 0$:

$$mgh = \tfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{2gh} = \sqrt{2gL(1-\cos\theta)}.$$

Numerically, $h = 1.5(1-\cos 60^\circ) = 1.5(1-0.5) = 0.75\ \text{m}$, so

$$v = \sqrt{2(9.8)(0.75)} = \sqrt{14.7} \approx 3.8\ \text{m/s}.$$

The same $v = \sqrt{2gh}$ result holds for a block sliding down any frictionless ramp or curved track through the same height drop, regardless of the shape of the path — only the vertical drop $h$ matters.

If nonconservative forces such as kinetic friction, air drag, or applied pushes do work, then

$$K_i + U_i + W_{\text{nc}} = K_f + U_f.$$

Equivalently,

$$W_{\text{nc}} = \Delta E_{\text{mech}}.$$

Friction usually decreases mechanical energy and converts it into thermal energy, so $W_f$ is usually negative for a sliding object.

Tip

Tip

If you forgot the sign of a non-conservative force, just think about whether the force would add energy or remove energy from the system (e.g. friction releases heat, meaning it removes energy).

Mechanical energy versus total energy

It is worth being careful about two different “totals.” Total mechanical energy $E_{\text{mech}} = K + U$ counts only kinetic and potential energy, and it is not conserved when nonconservative forces act — friction, drag, and inelastic deformation all lessen it away. Total energy, however, is always conserved: the mechanical energy lost to friction does not vanish, it reappears as thermal energy (and a little sound). If we write

$$\Delta E_{\text{mech}} + \Delta E_{\text{thermal}} + \cdots = 0,$$

energy is conserved overall, with any mechanical energy lost usually being converted to heat or sound. This ensures that we don’t violate the Law of Conservation of Energy.

Example. A block of mass $m = 3.0\ \text{kg}$ is released from rest and slides a distance $d = 4.0\ \text{m}$ down a $30^\circ$ incline with coefficient of kinetic friction $\mu_k = 0.20$. Find its speed at the bottom of that slide. Use $g = 9.8\ \text{m/s}^2$.

Gravity (conservative) and friction (nonconservative) both do work. Use $K_i + U_i + W_{\text{nc}} = K_f + U_f$ with the bottom of the slide as $U = 0$. The block drops a height $h = d\sin\theta$, so $U_i = mgd\sin\theta$. The normal force is $F_N = mg\cos\theta$, so friction does work

$$W_{\text{nc}} = -f_k d = -\mu_k mg\cos\theta\,d.$$

With $K_i = 0$:

$$mgd\sin\theta - \mu_k mg\cos\theta\,d = \tfrac{1}{2}mv^2.$$

Mass cancels, and solving for $v$:

$$v = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)}.$$

Plugging in, $\sin 30^\circ = 0.500$ and $\cos 30^\circ = 0.866$:

$$v = \sqrt{2(9.8)(4.0)\big(0.500 - 0.20(0.866)\big)} = \sqrt{78.4(0.327)} = \sqrt{25.6} \approx 5.1\ \text{m/s}.$$

For comparison, a frictionless incline would give $v = \sqrt{2gd\sin\theta} = \sqrt{39.2}\approx 6.3\ \text{m/s}$; friction has carried away the difference as heat.

Example. A $0.50\ \text{kg}$ block is pressed against a spring ($k = 800\ \text{N/m}$) compressed by $x = 0.12\ \text{m}$ on a horizontal surface. After release, the block crosses a rough patch of length $L = 1.2\ \text{m}$ with $\mu_k=0.25$, then climbs a frictionless ramp that rises by height $h=0.40\ \text{m}$. Find whether the block reaches the top of the ramp, and if it does, find its speed there.

The spring force is conservative, so its stored energy $U_s = \tfrac{1}{2}kx^2$ is the initial energy. Friction is the only nonconservative force. Apply

$$K_i + U_i + W_{\text{nc}} = K_f + U_f.$$

Here $K_i=0$ and the initial spring energy is $U_s=\tfrac12kx^2$. Friction removes $\mu_kmgL$ on the rough patch, and climbing the ramp requires gravitational potential energy $mgh$. If the remaining energy is positive, the block reaches the top:

$$\tfrac{1}{2}kx^2-\mu_kmgL-mgh=\tfrac{1}{2}mv^2.$$

Compute each energy term:

$$\tfrac{1}{2}kx^2=\tfrac{1}{2}(800)(0.12)^2=5.76\ \text{J},$$ $$\mu_kmgL=(0.25)(0.50)(9.8)(1.2)=1.47\ \text{J},$$

and

$$mgh=(0.50)(9.8)(0.40)=1.96\ \text{J}.$$

The remaining kinetic energy is

$$\tfrac12mv^2=5.76-1.47-1.96=2.33\ \text{J}.$$

Since this is positive, the block reaches the top. Its speed there is

$$v=\sqrt{\frac{2(2.33)}{0.50}}=\sqrt{9.32}\approx3.1\ \text{m/s}.$$

The clean strategy: spring energy in, friction and gravitational potential out, kinetic energy is whatever remains.

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Energy Diagrams and Equilibrium

In one-dimensional systems, a graph of $U(x)$ contains lots of useful information about force and an object’s current state. As a reminder, $F_x = -\frac{dU}{dx}.$ Equilibrium occurs where

$$\frac{dU}{dx}=0,$$

and at that point, the object has zero acceleration (since force is zero). The equilibrium is stable if $U(x)$ has a local minimum ($\frac{d^2 U}{dx^2} > 0$), unstable if it has a local maximum ($\frac{d^2 U}{dx^2} < 0$), and neutral if small displacements do not change $U$ to second order ($\frac{d^2 U}{dx^2} = 0$). A metastable (neutral) equilibrium is a local minimum that is stable for small disturbances but can escape if the total energy is high enough to cross a nearby barrier.

Example. A particle of mass $m = 0.20\ \text{kg}$ moves in one dimension under the potential

$$U(x) = 2x^2 - x^4$$

(in joules, with $x$ in meters). Find the equilibrium positions and classify them, find the force at $x = 0.5\ \text{m}$, and if the particle has total energy $E = 0.5\ \text{J}$ and is at $x=0$, find its speed there.

Equilibria. Set $dU/dx = 0$:

$$\frac{dU}{dx} = 4x - 4x^3 = 4x(1 - x^2) = 0 \;\Rightarrow\; x = 0,\ \pm 1.$$

To classify, use the second derivative $U''(x) = 4 - 12x^2$:

• At $x = 0$: $U'' = 4 > 0$, a local minimum → stable equilibrium.
• At $x = \pm 1$: $U'' = 4 - 12 = -8 < 0$, local maxima → unstable equilibria.

Force at $x = 0.5$. The force is

$$F_x = -\frac{dU}{dx} = -(4x - 4x^3) = -4(0.5) + 4(0.5)^3 = -2 + 0.5 = -1.5\ \text{N}.$$

The force points toward $-x$, i.e. back toward the stable minimum at the origin — a restoring force, as expected near a potential well.

Speed at $x = 0$. Here $U(0) = 0$, so all the energy is kinetic:

$$\tfrac{1}{2}mv^2 = E - U(0) = 0.5\ \text{J} \;\Rightarrow\; v = \sqrt{\frac{2(0.5)}{0.20}} = \sqrt{5} \approx 2.2\ \text{m/s}.$$

The particle is trapped in the well as long as $E$ is below the barrier height $U(\pm 1) = 2 - 1 = 1\ \text{J}$; its turning points are where $U(x) = E$. With $E = 0.5\ \text{J} < 1\ \text{J}$, it oscillates back and forth inside the well.

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Power

Power is defined as the rate of energy transfer:

$$P = \frac{dW}{dt}.$$

For a constant force acting on an object with instantaneous velocity $\vec{v}$,

$$P = \vec{F}\cdot \vec{v}.$$

Average power over a time interval is also defined as

$$\bar{P} = \frac{\Delta E}{\Delta t}.$$

Power is not a new kind of energy; it is how quickly energy is transferred or transformed.

Example. A car of mass $m = 1200\ \text{kg}$ drives up a $5.0^\circ$ incline at a constant $v = 25\ \text{m/s}$. Neglecting friction and drag, what power must the engine deliver? Use $g = 9.8\ \text{m/s}^2$.

At constant speed there is no change in kinetic energy, so the engine’s drive force must exactly balance the component of gravity along the incline:

$$F = mg\sin\theta = (1200)(9.8)\sin 5.0^\circ = (11760)(0.0872) \approx 1.03\times 10^3\ \text{N}.$$

Since the drive force is along the velocity, $P = Fv$:

$$P = (1.03\times 10^3\ \text{N})(25\ \text{m/s}) \approx 2.6\times 10^4\ \text{W} \approx 26\ \text{kW}.$$

Equivalently, the engine supplies gravitational potential energy at the rate $P = mg\,v\sin\theta = mg\,v_y$, where $v_y = v\sin\theta$ is the rate of gain of height. Both routes give the same answer because $P = \vec{F}\cdot\vec{v}$ counts only the force component along the motion.

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Key equations

Idea	Equation
Work, constant force	$W = \vec{F}\cdot\Delta\vec{r} = F\Delta r\cos\theta$
Work, variable force	$W = \int_{x_i}^{x_f} F_x\,dx$
Kinetic energy	$K = \tfrac{1}{2}mv^2$
Work-energy theorem	$W_{\text{net}} = \Delta K$
Conservative force / potential	$F_x = -\dfrac{dU}{dx}$
Gravity near Earth	$U_g = mgy$
Universal gravitation	$U_g = -\dfrac{GMm}{r}$
Spring potential energy	$U_s = \tfrac{1}{2}kx^2$
Mechanical energy conservation	$K_i + U_i = K_f + U_f$
With nonconservative work	$K_i + U_i + W_{\text{nc}} = K_f + U_f$
Escape speed	$v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}}$
Power	$P = \dfrac{dW}{dt} = \vec{F}\cdot\vec{v}$

Practice

Multiple Choice

1. A force is always perpendicular to a particle’s velocity. The force can change the particle’s

(A) speed but not direction

(B) direction but not speed

(C) kinetic energy only

(D) total mechanical energy only

2. A block slides up a rough incline and comes momentarily to rest. Compared with its mechanical energy at launch, its mechanical energy at the top is

(A) greater

(B) smaller

(C) the same

(D) zero

3. A force $F(x)=3x^2-2x$ acts on a particle from $x=0$ to $x=L$. The work done is

(A) $L^3-L^2$

(B) $3L^2-2L$

(C) $L^3+L^2$

(D) $3L^3-L^2$

4. If $U(x)=ax^4-bx^2$ with $a,b>0$, the force is

(A) $F_x=4ax^3-2bx$

(B) $F_x=-4ax^3+2bx$

(C) $F_x=ax^4-bx^2$

(D) $F_x=-a/x^4+b/x^2$

5. A particle moves in one dimension with potential energy $U(x)$. At a stable equilibrium,

(A) $U'=0$ and $U''>0$

(B) $U'=0$ and $U''<0$

(C) $U'>0$ and $U''=0$

(D) $U<0$ only

6. A block starts from rest at height $H$ above a horizontal spring, slides on a frictionless track, and compresses the spring a distance $x$. If the block instead starts from height $4H$, the new maximum compression is

(A) $x/2$

(B) $x$

(C) $2x$

(D) $4x$

7. A spring with constant $k$ is cut into two equal halves. One half is used as a spring. Compared with the original spring, the energy stored for the same stretch $x$ is

(A) half as large

(B) the same

(C) twice as large

(D) four times as large

8. A block moves through a region where a force $F(x)=F_0e^{-x/L}$ acts in the direction of motion. The work done from $x=0$ to $x=2L$ is

(A) $F_0L(1-e^{-2})$

(B) $2F_0L$

(C) $F_0L e^{-2}$

(D) $F_0/L$

9. A cart of mass $m$ moves under constant power $P$ from rest, with no resistive forces. Its speed after time $t$ is

(A) $Pt/m$

(B) $\sqrt{2Pt/m}$

(C) $2Pt/m$

(D) $\sqrt{Pt/(2m)}$

10. A satellite moves outward from radius $r$ to radius $2r$ around a planet of mass $M$. The work done by gravity during this motion is

(A) $-\dfrac{GMm}{2r}$

(B) $-\dfrac{GMm}{r}$

(C) $\dfrac{GMm}{2r}$

(D) zero, because gravity is perpendicular to orbital motion

11. A particle in potential $U(x)=\dfrac{A}{x^2}-\dfrac{B}{x}$, with $A,B>0$, has a stable equilibrium at

(A) $x=A/B$

(B) $x=2A/B$

(C) $x=B/A$

(D) $x=\sqrt{A/B}$

12. A projectile is launched upward from the surface of a planet of radius $R$ with speed $v_{\text{esc}}/2$. Neglect air resistance. Its maximum distance from the planet’s center is

(A) $4R/3$

(B) $3R/2$

(C) $2R$

(D) $4R$

FRQ

1. A block of mass $m$ starts from rest at height $H$ on a frictionless curved track, then crosses a rough horizontal patch of length $L$ with coefficient of kinetic friction $\mu_k$ before compressing a spring of constant $k$.

   $(A)$ Derive the speed of the block just before the rough patch.

   $(B)$ Determine the speed just after the rough patch.

   $(C)$ Find the maximum spring compression.

   $(D)$ Determine the condition on $H$ for the block to reach the spring.

2. A particle of mass $m$ moves in the potential $U(x)=ax^4-bx^2$, where $a,b>0$

   $(A)$ Find all equilibrium positions.

   $(B)$ Classify each equilibrium as stable or unstable.

   $(C)$ If the particle has total energy $E=0$, find its turning points.

   $(D)$ If the particle has total energy $E=0$, determine where its speed is greatest and justify your answer using the energy diagram.

3. A small spacecraft of mass $m$ moves radially away from a planet of mass $M$. Its engine supplies constant power $P$ for time $t_0$, starting from rest at radius $R$. Ignore air resistance and the changing mass of the spacecraft.

   $(A)$ Write an energy equation relating the spacecraft’s speed and radius after the burn.

   $(B)$ Determine the minimum engine energy needed for escape if the burn ends at radius $r_f$.

   $(C)$ Explain whether delivering the same energy quickly or slowly changes the escape condition in this idealized model.

   $(D)$ Identify one assumption in the model that would fail for a real rocket.