Chapter 5 · Work, Energy and Power

5.2 Potential Energy and Conservation of Mechanical Energy

Conservative forces such as gravity and the spring force store work as potential energy: $U=mgh$, $U=\tfrac12kx^2$. When only they do work, the mechanical energy $K+U$ never changes.

A roller-coaster car is hauled slowly to the top of the first hill and released. From then on there is no motor: the car plunges, races through the valley, climbs the next hill, slows, plunges again. Its speed changes enormously, yet the designers can tell you its speed at every point of the track before it is built — and they know that no later hill may be as high as the first. Something is being traded back and forth between height and speed, and the total is (almost) fixed.

In Section 5.1 you saw that the work done by gravity on a sliding box depends only on how far it drops, not on the slope. Forces with this property can "store" the work done against them and give it back later. The stored work is potential energy. Adding it to kinetic energy gives the mechanical energy, and when only such forces do work, mechanical energy is conserved. This one principle answers questions that would be very hard with Newton's laws alone: the speed of a pendulum bob, the height a spring launcher can reach, the speed a bob needs to loop a vertical circle. When friction acts, the same bookkeeping tells you exactly how much energy it has turned into heat.

Conservative forces and potential energy

Carry a $10\ \text{kg}$ box from the ground floor to a first-floor room $3.0\ \text{m}$ higher — up the stairs, in a lift, or by a long ramp. Gravity does $-mgh=-294\ \text{J}$ of work on it every time; bring it back down by any route and gravity does $+294\ \text{J}$. Over the round trip gravity's work is zero. Friction behaves quite differently: drag the box across a room and back, and friction does negative work on both legs; the longer the route, the more negative.

📘 Definition 5.4 — Conservative and non-conservative forces

A force is conservative if the work it does on a body moving between two points depends only on those end points, not on the path taken. Equivalently, the work it does around any closed path (ending where it started) is zero. Gravity and the force of an ideal spring are conservative.

A force that does not have this property is non-conservative. Kinetic friction and air resistance are non-conservative: the work they do depends on the length of the path and is negative around any closed path.

For a conservative force we can give every position a number — the potential energy — such that the work done by the force between two positions is simply the drop in that number.

📘 Definition 5.5 — Potential energy

For a conservative force, the change in potential energy $U$ of a body between two positions is minus the work done by the force:

$$\Delta U=U_f-U_i=-W_\text{c}.$$

Only changes in $U$ have a meaning. We are free to choose a reference position where $U=0$ (the floor, the table top, the natural length of a spring); changing the reference adds the same constant to every value of $U$ and changes no answer.

When a conservative force does positive work (a ball falling, a compressed spring pushing out), the potential energy decreases; when we do work against it (lifting the ball, compressing the spring), the potential energy increases by exactly that work. Reading Definition 5.5 for a small step $\Delta x$ along a line, $\Delta U=-F\,\Delta x$, so the force is minus the slope of the potential-energy graph.

📐 Result 5.4 — Force from potential energy (one dimension)

For a conservative force $F(x)$ acting along the $x$-axis, with potential energy $U(x)$,

$$U(x_f)-U(x_i)=-\int_{x_i}^{x_f}F(x)\,dx,$$ $$F(x)=-\frac{dU}{dx}.$$

The force points "downhill" on the $U$–$x$ graph — towards lower potential energy — and is zero where the graph is flat.

Gravitational and elastic potential energy

Gravity near the Earth. Take $y$ vertically up. The weight $-mg\,\hat{\mathbf{j}}$ is constant, so as a body rises by $h$ gravity does work $-mgh$ whatever the path (only the vertical part of the displacement counts). By Definition 5.5 its potential energy rises by $mgh$.

📐 Result 5.5 — Gravitational potential energy near the Earth

A body of mass $m$ at height $h$ above a chosen reference level has gravitational potential energy

$$U=mgh .$$

This holds when $g$ can be taken as constant — heights small compared with the Earth's radius (Chapter 7 treats large heights). Check: $F=-\dfrac{dU}{dh}=-mg$, the weight, pointing down ✓.

✏️ Example 1 — Path and reference do not matter
A passenger lifts a $15\ \text{kg}$ suitcase from the floor of a train onto a luggage rack $2.0\ \text{m}$ above the floor. (a) Find the increase in its gravitational potential energy. (b) Would the answer change if she first lifted it onto a seat and then onto the rack? (c) Taking the rack as the reference level, what is the suitcase's potential energy on the floor and on the rack?
(a) $\Delta U=mgh=15\times9.8\times2.0=294\ \text{J}\approx2.9\times10^2\ \text{J}$.
(b) No. Gravity is conservative: the work it does, and so $\Delta U$, depends only on the start and end heights.
(c) With $U=0$ at the rack: on the floor $h=-2.0\ \text{m}$, so $U=-294\ \text{J}$; on the rack $U=0$. The change is still $0-(-294)=+294\ \text{J}$.
Check A negative potential energy is perfectly allowed — it just means "below the reference level" ✓.

The spring force. Stretch or compress an ideal spring by $x$ from its natural length and it pulls or pushes back with a force proportional to $x$.

📘 Definition 5.6 — Spring force and spring constant

An ideal spring obeys Hooke's law: when its free end is displaced by $x$ from the natural (unstretched) length,

$$F_s=-kx ,$$

where the spring constant (force constant) $k$, in $\text{N/m}$, measures the stiffness. The minus sign means the force is restoring: it always points back towards $x=0$. Real springs obey this only for small enough $x$ (the elastic limit, Chapter 8).

The $F_s$–$x$ graph is a straight line through the origin with slope $-k$. As the end of the spring moves from $x_i$ to $x_f$, the spring's work is the signed area of a triangle-minus-triangle: $W_s=-\left(\tfrac12kx_f^2-\tfrac12kx_i^2\right)$. It depends only on the end points — the spring force is conservative — and comparing with $\Delta U=-W_s$ gives its potential energy.

📐 Result 5.6 — Elastic potential energy of a spring

An ideal spring of spring constant $k$, stretched or compressed by $x$ from its natural length, stores potential energy

$$U=\tfrac12kx^2 ,$$

with $U=0$ at the natural length. Check: $F_s=-\dfrac{dU}{dx}=-kx$ ✓. Since $U\propto x^2$, doubling the stretch stores four times the energy.

✏️ Example 2 — Energy in a compressed spring
A spring with $k=800\ \text{N/m}$ is compressed by $5.0\ \text{cm}$. (a) What force holds it there? (b) How much energy does it store? (c) The spring is allowed to extend until it is compressed by only $2.0\ \text{cm}$. How much work does the spring do on the block it pushes?
(a) $|F|=kx=800\times0.050=40\ \text{N}$.
(b) $U=\tfrac12kx^2=\tfrac12\times800\times(0.050)^2=1.0\ \text{J}$.
(c) At $2.0\ \text{cm}$: $U=\tfrac12\times800\times(0.020)^2=0.16\ \text{J}$. $W_s=-\Delta U=1.0-0.16=0.84\ \text{J}$ (positive: the spring pushes the way the block moves).
Check Area under the $F$–$x$ line from $2$ to $5\ \text{cm}$ is a trapezium: $\tfrac12(16+40)(0.030)=0.84\ \text{J}$ ✓. Note: the centimetres must be converted to metres before squaring.

Conservation of mechanical energy

📘 Definition 5.7 — Mechanical energy

The mechanical energy of a body (or system) is the sum of its kinetic and potential energies, $E=K+U$, where $U$ includes every conservative force that acts (gravity, springs).

Suppose only conservative forces do work. The work–energy theorem (Result 5.2) gives $W_\text{c}=\Delta K$; the definition of potential energy gives $W_\text{c}=-\Delta U$. So $\Delta K=-\Delta U$, i.e. $\Delta K+\Delta U=0$: whatever kinetic energy is gained is exactly the potential energy lost.

📐 Principle 5.7 — Conservation of mechanical energy

If only conservative forces do work on a body (other forces, such as a normal force or a string tension perpendicular to the motion, may act but do no work), its mechanical energy is constant:

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

If non-conservative forces (friction, air resistance, a push) also do work $W_\text{nc}$, then

$$W_\text{nc}=\Delta K+\Delta U=E_f-E_i .$$

When friction acts, $W_\text{nc}$ is negative and mechanical energy decreases — but it is not destroyed. It becomes thermal (internal) energy of the surfaces, sound and so on. Energy also exists as heat, chemical, electrical and nuclear energy, and in all known processes the total energy of an isolated system stays constant; mechanical energy is the part we can track with $K$ and $U$.

🧭 Problem-solving strategy — energy conservation
  1. Choose the system and two instants ("before" and "after"). Choose a zero level for gravitational $U$ (often the lowest point) and use $U=0$ at a spring's natural length.
  2. List the forces. Do any non-conservative forces do work? If none, $E_i=E_f$; if some, include $W_\text{nc}$.
  3. Write $\tfrac12mu^2+mgh_i+\tfrac12kx_i^2\ (+W_\text{nc})$ $=\tfrac12mv^2+mgh_f+\tfrac12kx_f^2$ and solve.
  4. If you need a force (a tension, a normal force), combine the speed from energy with Newton's second law at that point — energy alone never gives a force.
✏️ Example 3 — A playground slide
A $30\ \text{kg}$ child starts from rest at the top of a slide $4.0\ \text{m}$ high. (a) If the slide were frictionless, what would her speed be at the bottom? Does the shape of the slide matter? (b) She actually reaches the bottom at $7.0\ \text{m/s}$. How much mechanical energy was dissipated? (c) If the slide is $8.0\ \text{m}$ long, find the average friction force.
(a) Only gravity does work (the normal force is perpendicular to the motion). $mgh=\tfrac12mv^2$: $v=\sqrt{2gh}=\sqrt{2\times9.8\times4.0}=8.9\ \text{m/s}$. The shape does not matter, and neither does her mass.
(b) $E_i=mgh=30\times9.8\times4.0=1176\ \text{J}$; $\ E_f=\tfrac12\times30\times7.0^2=735\ \text{J}$. Dissipated: $441\ \text{J}\approx4.4\times10^2\ \text{J}$.
(c) $W_\text{nc}=-\bar f\times8.0=-441\ \text{J}$, so $\bar f=55\ \text{N}$.
Check Without friction she would keep all $1176\ \text{J}$; $\tfrac12(30)(8.85)^2=1175\ \text{J}$ ✓.
✏️ Example 4 — A spring launcher
A toy launcher has a spring with $k=400\ \text{N/m}$, compressed by $10\ \text{cm}$, behind a $50\ \text{g}$ ball. Neglect friction and air resistance. (a) Fired horizontally, how fast does the ball leave the spring? (b) Fired vertically upward, how high does the ball rise above its starting (compressed) position?
Stored $U_s=\tfrac12(400)(0.10)^2=2.0\ \text{J}$.
(a) Horizontal: no change of height. $\tfrac12mv^2=2.0$: $v=\sqrt{\dfrac{2\times2.0}{0.050}}=8.9\ \text{m/s}$.
(b) Vertical: at the top $K=0$ and the spring is relaxed, so all $2.0\ \text{J}$ has become $mgh$: $h=\dfrac{2.0}{0.050\times9.8}=4.1\ \text{m}$.
Check Measuring $h$ from the compressed position includes the $0.10\ \text{m}$ the ball rises while the spring extends — the energy method takes care of this automatically ✓.
Interactive 1 — Roller-Coaster Energy Bars

A $500\ \text{kg}$ car is released from rest $20\ \text{m}$ above the lowest point of the track. The bars show its kinetic energy $K$ (blue), gravitational potential energy $U=mgh$ (green), the energy turned into heat and sound by friction and drag (red), and the total (purple). With no resistance the car swings back and forth for ever and never rises above the dashed line; add resistance (modelled as a constant force, a fraction of the car's weight, opposing the motion) and watch the red bar eat into $K+U$ until the car settles in a valley.

0
height $h$—
speed $v$—
$K+U$—
dissipated—

Motion in a vertical circle. A bob on a string swung round a vertical circle is a classic energy problem. The tension is always along the radius, perpendicular to the velocity, so it does no work: mechanical energy is conserved. But the bob must also obey Newton's second law towards the centre (Section 3.4 and Chapter 4): at the top, $T+mg=\dfrac{mv^2}{L}$. The string stays taut only if $T\ge0$, so the speed at the top must be at least $\sqrt{gL}$.

✏️ Example 5 — Looping a vertical circle
A $100\ \text{g}$ bob hangs on a light string of length $L=0.80\ \text{m}$. It is given a horizontal speed $v_0$ at the lowest point. (a) Find the least $v_0$ for which it goes right round the vertical circle with the string taut. (b) For that $v_0$, find its speed at the top and the tension at the bottom.
Top Taut string: $T_\text{top}\ge0$, so $\dfrac{mv_\text{top}^2}{L}\ge mg$, i.e. $v_\text{top}^2\ge gL$.
Energy From bottom to top the bob rises $2L$: $\tfrac12mv_0^2=\tfrac12mv_\text{top}^2+mg(2L)$, so $v_0^2=v_\text{top}^2+4gL\ge5gL$.
(a) $v_0=\sqrt{5gL}=\sqrt{5\times9.8\times0.80}=6.3\ \text{m/s}$.
(b) $v_\text{top}=\sqrt{gL}=2.8\ \text{m/s}$. At the bottom, $T-mg=\dfrac{mv_0^2}{L}=5mg$, so $T=6mg=6\times0.100\times9.8=5.9\ \text{N}$.
Check $K_\text{bottom}:K_\text{top}=v_0^2:v_\text{top}^2=5:1$, and the difference $4mgL$ is the potential energy gained ✓. A step-by-step numerical simulation of the swing confirms that the tension just reaches zero at the top.

Energy of a block on a spring

For a block on a frictionless floor attached to a spring, $U=\tfrac12kx^2$ is a parabola. Pull the block out to $x=A$ and release it: $E=\tfrac12kA^2$. On a graph of energy against $x$, draw $E$ as a horizontal line: at each position the parabola gives $U$ and the gap up to the line is $K=E-U$. The block is fastest at $x=0$, where all the energy is kinetic, and stops and turns back at $x=\pm A$, where $U=E$ (the turning points); at every point $\tfrac12mv^2+\tfrac12kx^2=\tfrac12kA^2$.

✏️ Example 6 — The spring–block energy diagram
A $0.50\ \text{kg}$ block on a frictionless floor is attached to a spring with $k=200\ \text{N/m}$. It is pulled $10\ \text{cm}$ from the natural length and released. Find (a) its mechanical energy, (b) its greatest speed and where it occurs, (c) its speed when $x=6.0\ \text{cm}$, (d) its turning points.
(a) Released from rest at $x=A=0.10\ \text{m}$: $E=\tfrac12kA^2=\tfrac12(200)(0.10)^2=1.0\ \text{J}$.
(b) At $x=0$, $U=0$ and $K=E$: $v_\text{max}=\sqrt{\dfrac{2E}{m}}=\sqrt{\dfrac{2.0}{0.50}}=2.0\ \text{m/s}$.
(c) $U=\tfrac12(200)(0.060)^2=0.36\ \text{J}$, $K=0.64\ \text{J}$, $v=\sqrt{\dfrac{2\times0.64}{0.50}}=1.6\ \text{m/s}$.
(d) $U=E$ at $x=\pm0.10\ \text{m}$: it oscillates between $-10\ \text{cm}$ and $+10\ \text{cm}$.
Check These are the default settings of Interactive 2: the readouts show $E=1.00\ \text{J}$ and $v_\text{max}=2.00\ \text{m/s}$ ✓. (Chapter 13 studies this oscillation in detail.)
Interactive 2 — Spring, Block and the Potential-Energy Curve

A $0.50\ \text{kg}$ block on a frictionless floor is pulled out to $x=A$ and released; its motion is computed step by step from $F=-kx$. Below, the parabola is $U=\tfrac12kx^2$ and the dashed purple line is the total energy $E$. At the block's position the green segment is $U$ and the blue segment is $K=E-U$. The block turns round exactly where the line meets the curve.

200 N/m10 cm
total energy $E$—
$U$ now—
$K$ now—
greatest speed—
⚠️ Common mistakes

1. Using conservation when friction does work. $K+U=\text{constant}$ needs no non-conservative work. With friction or a motor, use $W_\text{nc}=E_f-E_i$.

2. Forgetting a potential energy. A block dropped onto a spring has both gravitational and spring potential energy changing; include every conservative force.

3. Centimetres in $\tfrac12kx^2$. Convert to metres before squaring: $5\ \text{cm}$ is $0.05\ \text{m}$, and $0.05^2=0.0025$, not $25$.

4. "Speed at the top of a vertical circle can be zero." For a bob on a string it must be at least $\sqrt{gL}$; for a bead threaded on a rigid ring (which can push as well as pull) it can be zero. The condition comes from Newton's second law, not from energy.

5. Potential energy of what? Strictly, $mgh$ belongs to the body–Earth system and $\tfrac12kx^2$ to the spring. Saying "the ball's potential energy" is a convenient shorthand.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced) — potential-energy curves; equilibrium and stability

Not in the CBSE syllabus or the NCERT chapter, and not named in the JEE Main, JEE Advanced or NEET 2026 syllabi — but it follows directly from $F=-\dfrac{dU}{dx}$ and is regularly set in JEE Main and JEE Advanced papers.

Reading any $U$–$x$ curve. Draw the total energy $E$ as a horizontal line. Since $K=E-U$ can never be negative, the body can only be where $U\le E$; where the curve meets the line it stops and turns back — just as the block on the spring does at $x=\pm A$.

A body is in equilibrium ($F=0$) wherever the $U$–$x$ graph is flat.

Stable equilibrium: a minimum of $U$ — displace the body slightly and the force pushes it back (a ball at the bottom of a bowl, the block at $x=0$ on a spring). Unstable: a maximum of $U$ — a small displacement produces a force pushing it further away (a ball balanced on a dome). Neutral: $U$ constant nearby — no force after a displacement (a ball on a level floor). A test: $\dfrac{d^2U}{dx^2}\gt0$ means stable, $\lt0$ unstable.

Example. $U(x)=(2x^3-6x)\ \text{J}$ ($x$ in m). Then $F=-\dfrac{dU}{dx}=6-6x^2$, zero at $x=\pm1\ \text{m}$. At $x=1\ \text{m}$, $U=-4\ \text{J}$ is a minimum (stable); at $x=-1\ \text{m}$, $U=+4\ \text{J}$ is a maximum (unstable). A particle resting at $x=1\ \text{m}$ needs more than $8\ \text{J}$ of kinetic energy to escape over the hump at $x=-1\ \text{m}$.

Exercises

Take $g=9.8\ \text{m/s}^2$ and neglect air resistance unless told otherwise.

I
Potential energy.
  • 1 A $2.0\ \text{kg}$ book is lifted from the floor to a shelf $1.5\ \text{m}$ high. Find the change in its gravitational potential energy.
  • 2 A $60\ \text{kg}$ trekker climbs $500\ \text{m}$ (vertically) up a winding mountain path. Find the increase in her potential energy. Would a steeper, shorter path change it?
  • 3 A $0.50\ \text{kg}$ ball rests on a table $0.80\ \text{m}$ high; a shelf is $2.0\ \text{m}$ above the floor. Find the ball's potential energy with the reference level at (a) the floor, (b) the table top, (c) the shelf. Find the change in $U$ when the ball falls to the floor, in each case.
  • 4 Find the energy stored in a spring with $k=500\ \text{N/m}$ stretched by $4.0\ \text{cm}$.
  • 5 A force of $30\ \text{N}$ stretches a spring by $6.0\ \text{cm}$. Find $k$ and the energy stored.
  • 6 How much more work is needed to stretch the spring of part 5 from $6.0\ \text{cm}$ to $12\ \text{cm}$?
  • 7 A spring stores energy $U$ when compressed by $x$. How much does it store when compressed by $3x$? When stretched by $x$?
  • 8 A conservative force has potential energy $U(x)=(5x^2-3x)\ \text{J}$, $x$ in metres. Find the force at $x=0.50\ \text{m}$, and the position where the force is zero.
1 $\Delta U=mgh=2.0\times9.8\times1.5=29\ \text{J}$.
2 $\Delta U=60\times9.8\times500=2.9\times10^5\ \text{J}$. No — gravity is conservative; only the height gained matters.
3 (a) $0.50\times9.8\times0.80=3.9\ \text{J}$; (b) $0$; (c) $0.50\times9.8\times(-1.2)=-5.9\ \text{J}$. The change on falling to the floor is $-3.9\ \text{J}$ in every case.
4 $U=\tfrac12(500)(0.040)^2=0.40\ \text{J}$.
5 $k=\dfrac{30}{0.060}=500\ \text{N/m}$; $U=\tfrac12kx^2=\tfrac12Fx=\tfrac12(30)(0.060)=0.90\ \text{J}$.
6 $\tfrac12(500)\left(0.12^2-0.060^2\right)=250\times0.0108=2.7\ \text{J}$ — three times the work of the first $6\ \text{cm}$.
7 $U\propto x^2$: $9U$. Stretched by $x$: also $U$ (same $x^2$).
8 $F=-\dfrac{dU}{dx}=-(10x-3)$; at $x=0.50\ \text{m}$, $F=-2.0\ \text{N}$ (along $-x$). $F=0$ at $x=0.30\ \text{m}$.
II
Conservation of mechanical energy.
  • 1 A stone is dropped from a cliff $45\ \text{m}$ high. Find its speed just before it hits the water.
  • 2 A ball is thrown straight up at $14\ \text{m/s}$. Use energy to find how high it rises.
  • 3 A ball is thrown at $10\ \text{m/s}$ from a balcony $25\ \text{m}$ above the ground. Find its speed on reaching the ground. Does the direction of the throw matter?
  • 4 A skateboarder moving at $5.0\ \text{m/s}$ at the bottom of a smooth ramp coasts up it. How high does she rise?
  • 5 A simple pendulum of length $1.0\ \text{m}$ is released from rest with the string at $60^\circ$ to the vertical. Find the bob's speed at the lowest point.
  • 6 For the pendulum of part 5, find the speed when the string makes $30^\circ$ with the vertical.
  • 7 A spring with $k=2000\ \text{N/m}$, compressed by $15\ \text{cm}$, launches a $0.40\ \text{kg}$ block along a frictionless horizontal surface. Find the block's speed.
  • 8 The block of part 7 then slides up a frictionless ramp. How high does it rise?
1 $v=\sqrt{2gh}=\sqrt{2\times9.8\times45}=29.7\ \text{m/s}\approx30\ \text{m/s}$.
2 $h=\dfrac{u^2}{2g}=\dfrac{196}{19.6}=10\ \text{m}$.
3 $v^2=u^2+2gh=100+490=590$, $v=24\ \text{m/s}$ (24.3). No: energy is a scalar, so the landing speed is the same in any direction (the landing velocity differs).
4 $h=\dfrac{v^2}{2g}=\dfrac{25}{19.6}=1.3\ \text{m}$.
5 Drop in height $L(1-\cos60^\circ)=0.50\ \text{m}$: $v=\sqrt{2\times9.8\times0.50}=3.1\ \text{m/s}$.
6 Drop from $60^\circ$ to $30^\circ$: $L(\cos30^\circ-\cos60^\circ)=0.366\ \text{m}$; $v=\sqrt{2\times9.8\times0.366}=2.7\ \text{m/s}$.
7 $U=\tfrac12(2000)(0.15)^2=22.5\ \text{J}$; $v=\sqrt{\dfrac{2\times22.5}{0.40}}=10.6\ \text{m/s}\approx11\ \text{m/s}$.
8 $h=\dfrac{22.5}{0.40\times9.8}=5.7\ \text{m}$.
III
When non-conservative forces act.
  • 1 A $40\ \text{kg}$ child slides from rest down a slide $3.0\ \text{m}$ high and reaches the bottom at $6.0\ \text{m/s}$. How much mechanical energy is dissipated?
  • 2 The slide in part 1 is $6.0\ \text{m}$ long. Find the average friction force.
  • 3 A $0.15\ \text{kg}$ ball dropped from $10\ \text{m}$ rebounds to $6.4\ \text{m}$. How much mechanical energy is lost in the bounce, and what percentage is that?
  • 4 A $2.0\ \text{kg}$ block sliding at $6.0\ \text{m/s}$ on a level floor with $\mu_k=0.30$ comes to rest. How far does it slide?
  • 5 A $1000\ \text{kg}$ car coasts from rest down a hill $20\ \text{m}$ high, engine off, and reaches the bottom at $15\ \text{m/s}$. Find the work done by the resistive forces.
  • 6 A $1.0\ \text{kg}$ block moving at $4.0\ \text{m/s}$ on a frictionless floor runs into a spring with $k=400\ \text{N/m}$. Find the greatest compression.
  • 7 Repeat part 6 if the floor has $\mu_k=0.20$ while the spring is being compressed (the block still meets the spring at $4.0\ \text{m/s}$).
1 $mgh=40\times9.8\times3.0=1176\ \text{J}$, $K=\tfrac12(40)(6.0)^2=720\ \text{J}$; dissipated $456\ \text{J}\approx4.6\times10^2\ \text{J}$.
2 $\bar f=\dfrac{456}{6.0}=76\ \text{N}$.
3 $\Delta E=mg(10-6.4)=0.15\times9.8\times3.6=5.3\ \text{J}$; that is $\tfrac{3.6}{10}=36\%$ of the original energy.
4 $\mu_kmg\,s=\tfrac12mv^2$: $s=\dfrac{v^2}{2\mu_kg}=\dfrac{36}{5.88}=6.1\ \text{m}$.
5 $W_\text{nc}=E_f-E_i=\tfrac12(1000)(15)^2-1000\times9.8\times20=112\,500-196\,000=-8.4\times10^4\ \text{J}$.
6 $\tfrac12kx^2=\tfrac12mv^2=8.0\ \text{J}$: $x=\sqrt{\dfrac{16}{400}}=0.20\ \text{m}$.
7 $8.0=\tfrac12(400)x^2+\mu_kmg\,x$: $200x^2+1.96x-8.0=0$, $x=\dfrac{-1.96+\sqrt{1.96^2+6400}}{400}=0.195\ \text{m}$ — slightly less, as friction takes some energy.
IV
Applications and harder problems.
  • 1 A bob on a $0.80\ \text{m}$ string is to go round a vertical circle. Find the least speed it needs at the bottom, and its speed at the top in that case.
  • 2 In part 1 (least speed), show that the tension at the bottom is six times the bob's weight.
  • 3 A small car runs without friction down a track and round a vertical loop of radius $R=4.0\ \text{m}$. From what least height above the bottom of the loop must it start from rest so as to stay on the track at the top? (At the top the track can only push down on the car.)
  • 4 A $0.20\ \text{kg}$ ball is dropped from $0.50\ \text{m}$ above the top of a vertical spring with $k=196\ \text{N/m}$. Find the greatest compression of the spring.
  • 5 For the block of Example 6, find its speed at $x=5.0\ \text{cm}$, and the position where its kinetic and potential energies are equal.
  • 6 (Modelling) A pole vaulter runs in at $9.5\ \text{m/s}$. If all her kinetic energy became gravitational potential energy, how high would her centre of mass rise? Her centre of mass starts about $1.0\ \text{m}$ above the ground; the best vaulters clear over $6\ \text{m}$. Suggest where the extra height comes from.
  • 7 Prove that a body sliding from rest down any frictionless track that descends a height $h$ reaches the bottom at speed $\sqrt{2gh}$. Why does the shape of the track not matter, although the time taken does?
1 $v_0=\sqrt{5gL}=\sqrt{39.2}=6.3\ \text{m/s}$; $v_\text{top}=\sqrt{gL}=2.8\ \text{m/s}$ (Example 5).
2 At the bottom $T-mg=\dfrac{mv_0^2}{L}=\dfrac{m(5gL)}{L}=5mg$, so $T=6mg$.
3 At the top, $mg+N=\dfrac{mv^2}{R}$ with $N\ge0$: $v_\text{top}^2\ge gR$. Energy: $mgh=mg(2R)+\tfrac12mv_\text{top}^2\ge2mgR+\tfrac12mgR$, so $h\ge2.5R=10\ \text{m}$.
4 Lowest point is $(0.50+x)$ below the start: $\tfrac12kx^2=mg(0.50+x)$, i.e. $98x^2-1.96x-0.98=0$, $x=\dfrac{1.96+\sqrt{1.96^2+4(98)(0.98)}}{196}=0.11\ \text{m}$.
5 $U=\tfrac12(200)(0.050)^2=0.25\ \text{J}$, $K=0.75\ \text{J}$, $v=\sqrt{3.0}=1.7\ \text{m/s}$. $K=U=\tfrac12E$ when $\tfrac12kx^2=\tfrac14kA^2$: $x=\pm\dfrac{A}{\sqrt2}=\pm7.1\ \text{cm}$.
6 $h=\dfrac{v^2}{2g}=\dfrac{90.25}{19.6}=4.6\ \text{m}$, so the centre of mass could reach about $5.6\ \text{m}$. The rest comes from work done by her arms and body pushing up on the pole during the vault (and a centre of mass that passes just over, or even under, the bar).
7 Only gravity does work (the normal force is perpendicular to the track everywhere), and gravity's work $mgh$ is path-independent: $\tfrac12mv^2=mgh$. The time depends on the acceleration along the way, which does depend on the shape.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 5 marks A body of mass $m$ is dropped from rest at a height $H$ above the ground. Show that its mechanical energy is the same (a) at the start, (b) after it has fallen a distance $x$, (c) just before it hits the ground. State the principle illustrated and its condition.

(a) $K=0$, $U=mgH$: $E=mgH$. [1]
(b) $v^2=2gx$ (Chapter 2), so $K=mgx$; height is $H-x$, so $U=mg(H-x)$. $E=mgx+mgH-mgx=mgH$. [1½]
(c) $v^2=2gH$: $K=mgH$, $U=0$, $E=mgH$. [1½]
Principle Conservation of mechanical energy: $K+U$ is constant when only conservative forces (here gravity) do work — air resistance neglected. [1]
Exam tip Draw a small table of $K$, $U$, $E$ at the three points — examiners like it, and it makes the constant total obvious.

E2. NEET Two springs A and B with $k_A=2k_B$ are stretched by the same force. The ratio of the energies stored, $U_A:U_B$, is (a) $2:1$ (b) $1:2$ (c) $4:1$ (d) $1:4$

Idea Same force: $x=\dfrac Fk$, so $U=\tfrac12kx^2=\dfrac{F^2}{2k}\propto\dfrac1k$.
Answer (b) $1:2$ — the stiffer spring stretches less and stores less.
Exam tip For the same extension the answer would be $2:1$ (a). Read whether force or extension is equal.

E3. JEE Main (Numerical value) A small bob on a light string of length $0.50\ \text{m}$ is given a horizontal speed at its lowest point. Taking $g=10\ \text{m/s}^2$, find the least speed (in m/s) for which it completes the vertical circle.

Condition $T_\text{top}\ge0$ gives $v_\text{top}^2\ge gL$; energy gives $v_0^2=v_\text{top}^2+4gL\ge5gL$.
Answer $v_0=\sqrt{5\times10\times0.50}=5\ \text{m/s}$.
Exam tip $\sqrt{4gL}$ (enough to reach the top with zero speed) is the trap: a string goes slack before then. It is the right answer only for a bob on a rigid rod.

E4. JEE Advanced (One or more correct) A block of mass $m$ is dropped from rest at a height $h$ above the top of a light vertical spring of constant $k$. Air resistance is negligible. Which statements are correct? (A) The block's speed is greatest when the spring is compressed by $mg/k$. (B) The greatest compression $x_m$ satisfies $\tfrac12kx_m^2=mg(h+x_m)$. (C) At the greatest compression the net force on the block is zero. (D) The greatest kinetic energy of the block is $mgh+\dfrac{m^2g^2}{2k}$.

(A) The block speeds up while $mg\gt kx$ and slows once $kx\gt mg$: fastest at $x_0=mg/k$ ✓.
(B) At $x_m$ the block is momentarily at rest, having fallen $h+x_m$: $mg(h+x_m)=\tfrac12kx_m^2$ ✓.
(C) At $x_m$, $kx_m\gt mg$: the net force is upward, not zero ✗ (zero net force is at $x_0$).
(D) $K_\text{max}=mg(h+x_0)-\tfrac12kx_0^2=mgh+\dfrac{m^2g^2}{k}-\dfrac{m^2g^2}{2k}=mgh+\dfrac{m^2g^2}{2k}$ ✓.
Answer (A), (B), (D). A numerical scan of $K(x)=mg(h+x)-\tfrac12kx^2$ confirms the maximum at $x_0$.

E5. JEE Main NEET The potential energy of a particle moving along the $x$-axis is $U=(4x^2-2x)\ \text{J}$, $x$ in metres. The force on it at $x=1\ \text{m}$ is (a) $6\ \text{N}$ along $+x$ (b) $6\ \text{N}$ along $-x$ (c) $2\ \text{N}$ along $-x$ (d) $8\ \text{N}$ along $+x$

Idea $F=-\dfrac{dU}{dx}=-(8x-2)$; at $x=1$, $F=-6\ \text{N}$.
Answer (b).
Exam tip (a) forgets the minus sign; (c) is $U(1)$ itself. The force points towards lower $U$.

E6. Board · Case study A $400\ \text{kg}$ roller-coaster car is released from rest at point A, $30\ \text{m}$ above the ground. The track dips to B, $5.0\ \text{m}$ above the ground, then climbs to the top of a hill C, $18\ \text{m}$ above the ground. (i) Neglecting friction, find the speed at B. [2] (ii) Neglecting friction, find the speed at C. [1] (iii) In fact the car reaches C at $14\ \text{m/s}$. How much energy was dissipated between A and C? [1]

(i) Drop $25\ \text{m}$: $v_B=\sqrt{2\times9.8\times25}=22\ \text{m/s}$ ($22.1$). [2]
(ii) Drop $12\ \text{m}$: $v_C=\sqrt{2\times9.8\times12}=15\ \text{m/s}$ ($15.3$). [1]
(iii) $mg(12)-\tfrac12m(14)^2=47\,040-39\,200=7.8\times10^3\ \text{J}$. [1]
Exam tip The mass cancels in (i) and (ii) but not in (iii) — energy dissipated is an amount of energy, so it needs $m$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much energy can a pumped-storage hydroelectric plant store as gravitational potential energy, and how much of it comes back?

Why it matters. Solar and wind power do not follow demand. Grids in India, China, the UK, Switzerland, the USA and elsewhere store surplus electricity by pumping water up into a high reservoir, then let it fall through turbines when power is needed. The stored energy is simply $mgh$ for the water — Result 5.5 on a huge scale — but pumps and turbines lose some energy each way.

What to do.

  1. An upper reservoir holds $2.0\times10^6\ \text{m}^3$ of water (density $1.0\times10^3\ \text{kg/m}^3$) at an average height of $300\ \text{m}$ above the turbines. Compute the stored energy in joules and in megawatt-hours ($1\ \text{MWh}=3.6\times10^9\ \text{J}$).
  2. If the round-trip efficiency is $75\%$, how much electrical energy is recovered? If it is delivered over $6$ hours, what is the average power output?
  3. How many homes using $10\ \text{kWh}$ a day could this supply for one day?
  4. Compare with a plant of the same volume but only $50\ \text{m}$ of head. Why do engineers seek mountain sites with large height differences?
  5. Find data for a real pumped-storage station and repeat your estimate. How close is it to the published capacity?

Tools: Calculator or spreadsheet; operators' and government energy-agency websites for real plant data.

Hand in: Your calculations, a short comparison table (high-head vs low-head), and a paragraph on why pumped storage is called a "water battery".

Going further. Where does the lost $25\%$ go? List the non-conservative forces and processes (friction in pipes, turbine and pump losses, electrical heating) and suggest which could be reduced.

Instructor note: $E=\rho Vgh=5.88\times10^{12}\ \text{J}\approx1.63\times10^3\ \text{MWh}$; at $75\%$, about $1.2\times10^3\ \text{MWh}$, i.e. about $200\ \text{MW}$ for $6\ \text{h}$ — enough for roughly $1.2\times10^5$ homes for a day. At $50\ \text{m}$ head the same volume stores one-sixth as much. Students should realise that $U\propto h$, so height is cheaper than reservoir volume.

🧪 Self-check quiz

0/5 correct

Q1. Which of these forces is non-conservative?

Q2. The extension of a spring is doubled. The elastic potential energy stored becomes

Q3. A ball is dropped from a height of $20\ \text{m}$. Taking $g=10\ \text{m/s}^2$ and neglecting air resistance, its speed on reaching the ground is

Q4. Identical blocks slide from rest down three frictionless ramps of the same height but different slopes. At the bottom,

Q5. In a region where a particle's potential energy $U$ increases as $x$ increases, the conservative force on it

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