PHY1001 · Week 13 · Potential Energy and Conservation of Energy: Conservation of Energy, Changes in Mechanical Energy, relationship Between Conservative Forces and Potential Energy, Mass-Energy Equivalence.

Conservation of mechanical energy

K + U is constant; with friction the loss is f·d. You solve free fall, pendulum, spring, ramp and loop questions with a single equation.

conservation of mechanical energymechanical energyisolated systeminternal energynonconservative workenergy diagramturning pointrest energymass-energy equivalence

The pieces of the last two weeks come together here. We call the sum of kinetic energy (week 11) and potential energy (week 12) the mechanical energy. If only conservative forces do work, this sum does not change. This is one of the most powerful shortcuts in physics: however complicated the shape of the ramp, you find the speed or height by looking only at the initial and final states.

Golden rules:

Conservation: $K_i + U_i = K_f + U_f$ , only if only conservative forces (gravity, spring) do work.
If there is friction, the mechanical energy decreases: $\Delta E_{mech} = -f_kd$ .
Energy is a scalar. No direction, no components. If the direction of the velocity is not asked, energy is the shortest way.
The normal force does no work and does not appear in the equation.

Conservation of mechanical energy

The work of a conservative force is $W = -\Delta U$ and the work-energy theorem is $W = \Delta K$ . Equating them:

$\Delta K + \Delta U = 0 \qquad\Longleftrightarrow\qquad K_i + U_i = K_f + U_f$

$\tfrac12 mv_i^2 + mgy_i + \tfrac12 kx_i^2 = \tfrac12 mv_f^2 + mgy_f + \tfrac12 kx_f^2$

Solution method

1. Choose the system and the two moments (initial / final).
2. Choose the reference level (y = 0), usually the lowest point.
3. Write K, U_g, U_s for each moment. Drop the ones that are zero.
4. Is there friction? If so, add -f d to the right side.
5. Solve for the unknown.

Worked example 1. A ball is thrown upward at 15.0 m/s from the edge of a roof 10.0 m high. What is its speed when it hits the ground?

Reference: ground.  y_i = 10.0 m, y_f = 0
½ v_i² + g y_i = ½ v_f²        (m cancelled)
v_f² = 15.0² + 2 · 9.80 · 10.0 = 225 + 196 = 421
v_f = 20.5 m/s

Whether the ball is thrown up, down or sideways changes nothing. Every ball thrown with the same speed hits the ground with the same speed. Energy does not know direction.

PITFALL · energy does not give direction. The conservation equation gives the speed. If the impact angle is asked, you go back to components and kinematics.

Choosing the system

Potential energy belongs not to a single object but to a system: gravitational potential energy to the object and the Earth, spring energy to the block and the spring. If only conservative forces do work inside the system and no work comes from outside, the system is considered isolated and its mechanical energy is conserved. On the exam think of it in one sentence: “Is there friction, air resistance or a push?” If not, conservation; if so, a loss term.

With a spring

Worked example 2. A 2.00 kg block rests against a spring with $k = 500\ \text{N/m}$ , and the spring is compressed 0.200 m. It is released, goes along a frictionless horizontal path and climbs a frictionless ramp. What are its speed when it leaves the spring and the maximum height it can reach?

U_s = ½ · 500 · 0.200² = 10.0 J
Leaving the spring: ½ · 2.00 · v² = 10.0  →  v = 3.16 m/s
At the highest point K = 0:  m g h = 10.0  →  h = 10.0 / (2.00 · 9.80) = 0.510 m

Pendulum

When the pendulum bob is at an angle $\theta$ to the vertical, its height above the lowest point is $h = L(1 - \cos\theta)$ . Since the string tension is perpendicular to the motion, it does no work. The speed at the bottom is $v = \sqrt{2gL(1 - \cos\theta)}$ .

With friction: change in mechanical energy

Kinetic friction is nonconservative. The work it does reduces the mechanical energy, and that energy goes into the internal energy of the surfaces, that is, heat:

$\Delta E_{mech} = \Delta K + \Delta U = -f_kd$

$d$ is the length of the sliding path. If an external force (you or a motor) does work, you add that too:

$K_f + U_f = K_i + U_i + W_{\text{ext}} - f_kd$

Worked example 3. A 5.00 kg box starts sliding on a horizontal floor at 8.00 m/s, $\mu_k = 0.250$ . How far does it go, and how many J go into internal energy?

In the final state K = 0, U does not change.
0 = ½ · 5.00 · 8.00² - 0.250 · 5.00 · 9.80 · d
160 J = 12.25 N · d  →  d = 13.1 m
Going into internal energy: 160 J (all the kinetic energy)

PITFALL · friction on an incline is not $\mu mg$ . On a ramp $N = mg\cos\theta$ , so $f_k = \mu_k mg\cos\theta$ . If you write $\mu_k mg$ out of the habit from the horizontal, you calculate too much friction.

Energy diagrams

Draw the total energy $E$ as a horizontal line over the $U(x)$ graph. At every point $K = E - U(x)$ :

$E > U$ : the particle can be there, $K > 0$ .
$E = U$ : a turning point. The particle stops momentarily and turns back.
$E < U$ : a forbidden region, $K$ cannot be negative.
At the minimum of $U$ , $K$ is largest (the place where it passes fastest). This is the stable equilibrium point.

For a block on a spring, $U = \tfrac12 kx^2$ is a parabola. The turning points are $x = \pm A$ (amplitude), and the fastest point is $x = 0$ .

TIP · the force is not zero at a turning point. The velocity is zero, but the force $-dU/dx$ is at one of its largest values. It is also this force that turns it back. Like the acceleration of a ball thrown upward being $a = -g$ at the top.

Mass-energy equivalence (short note)

According to Einstein, mass is also a form of energy. The rest energy of an object at rest:

$E_R = mc^2 \qquad c = 3.00\times10^8\ \text{m/s}$

Worked example 4. The energy equivalent of a 1.00 kg mass is $(1.00)(3.00\times10^8)^2 = 9.00\times10^{16}\ \text{J}$ . That is the electricity a big city would use for years. In nuclear reactions a small fraction of the mass is converted into energy this way. The trap on the exam is the unit: if the mass is given in grams, convert to kilograms.

Mixed exam examples

Worked example 5 (spring + friction). A 0.500 kg block is held against a spring with $k = 200\ \text{N/m}$ compressed 0.150 m. It is released and slides on a horizontal floor with $\mu_k = 0.300$ (friction also acts while the spring is extending). How far from the release point does the block stop?

Initial: K = 0, U_s = ½ · 200 · 0.150² = 2.25 J
Final:   K = 0, U_s = 0 (the block has left the spring)
Friction: f = 0.300 · 0.500 · 9.80 = 1.47 N
0 = 2.25 - 1.47 d   →   d = 1.53 m

Here the block leaves the spring after 0.150 m and travels a total of 1.53 m. All the energy of the spring turns into heat in the end.

Worked example 6 (normal force in a loop). A 0.200 kg car is released from rest at a height of $h = 3R = 1.50$ m above the bottom of a loop of radius $R = 0.500$ m. What are the speed and the normal force at the top of the loop?

Energy:  g h = g (2R) + ½ v²   →  v² = 2g(h - 2R) = 2·9.80·0.500 = 9.80
         v = 3.13 m/s
At the top (center below):  N + m g = m v² / R
         N = 0.200 (9.80/0.500 - 9.80) = 0.200 · 9.80 = 1.96 N

This question is a combination of the dynamics of circular motion of week 8 and energy. First find the speed with energy, then write Newton’s second law toward the center.

Situation	Equation
Only gravity and spring	$K_i + U_i = K_f + U_f$
Kinetic friction present	$K_f + U_f = K_i + U_i - f_kd$
An external push or motor present	$K_f + U_f = K_i + U_i + W_{\text{ext}} - f_kd$
A collision (sticking) present	At the moment of collision not energy but momentum (week 14)

Common mistakes

Wrong idea	The truth
$K_i + U_i = K_f + U_f$ with friction present	Add the loss $f_kd$ to one side
$v = \sqrt{gh}$	$v = \sqrt{2gh}$
$h = L$ or $L\cos\theta$ in a pendulum	$h = L(1 - \cos\theta)$
Forgetting the initial speed	Add $K_i = \tfrac12 mv_i^2$
Adding the speeds directly	Add the energies, then take the square root
$f = \mu mg$ on an incline	$f = \mu mg\cos\theta$
The energy of a spring is $kx$	$\tfrac12 kx^2$
The speed at the top of a loop can be zero	At least $v^2 = gR$ ; $h_{min} = 2.5R$
The force is zero at a turning point	$K = 0$ , the force is not zero
Using grams in $E = mc^2$	Convert to kg: 1 g = $10^{-3}$ kg
Energy gives direction too	Only speed; use components for direction

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

A stone is dropped from rest from a height of 12.0 m. Ignoring air resistance, what is its speed just before it hits the ground?

15.3 m/s
10.8 m/s
235 m/s
118 m/s
7.67 m/s

Show answer

A) 15.3 m/s

$mgh = \tfrac12 mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2(9.80)(12.0)} = 15.3\ \text{m/s}$ . B: 2 forgotten ( $\sqrt{gh}$ ). C: square root not taken ( $v^2$ ). D: $gh$ , with neither the 2 nor the root. E: $\sqrt{gh/2}$ , the $\tfrac12$ moved to the wrong side.

Question 2medium

A simple pendulum of length 1.20 m is released from rest at $40.0^\circ$ from the vertical. What is the bob's speed at the lowest point?

4.85 m/s
4.24 m/s
2.35 m/s
6.26 m/s
3.89 m/s

Show answer

C) 2.35 m/s

The vertical drop is $h = L(1 - \cos 40.0^\circ) = 1.20(0.234) = 0.281\ \text{m}$ . $v = \sqrt{2gh} = 2.35\ \text{m/s}$ . A: took $h = L$ . B: $h = L\cos\theta$ . D: the calculator is in RAD mode. E: $h = L\sin\theta$ .

Question 3medium

A spring ( $k = 600\ \text{N/m}$ ) standing vertically is compressed $0.150\ \text{m}$ and used to launch a 0.250 kg ball straight up. Ignoring air resistance and the spring's mass, how high above its release point does the ball rise?

5.51 m
36.7 m
2.61 m
2.76 m
1.38 m

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D) 2.76 m

$\tfrac12 kx^2 = mgh \Rightarrow h = \dfrac{\tfrac12(600)(0.150)^2}{(0.250)(9.80)} = \dfrac{6.75}{2.45} = 2.76\ \text{m}$ . A: $\tfrac12$ forgotten. B: $kx/mg$ , a ratio of forces. C: 0.150 m was subtracted; the question asks from the release point. E: an extra $\tfrac12$ .

Question 4medium

A block starts from rest and slides $4.00\ \text{m}$ down a $30.0^\circ$ incline with $\mu_k = 0.200$ . What is its speed at the bottom?

6.26 m/s
4.85 m/s
5.06 m/s
7.75 m/s
7.26 m/s

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C) 5.06 m/s

$\tfrac12 mv^2 = mgd\sin\theta - \mu_k mg\cos\theta\, d \Rightarrow v = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)} = 5.06\ \text{m/s}$ . A: frictionless. B: took $N = mg$ (cos forgotten). D: sin and cos swapped. E: the sign reversed as if friction added energy.

Question 5easy

A 5.00 kg crate slides $3.00\ \text{m}$ across a level floor with $\mu_k = 0.400$ . How much mechanical energy is converted into internal energy?

58.8 J
19.6 J
147 J
6.00 J
0 J

Show answer

A) 58.8 J

$\Delta E_{mech} = -f_kd = -\mu_k mg\,d = -(0.400)(5.00)(9.80)(3.00) = -58.8\ \text{J}$ . This much energy goes into internal energy (heat). B: the friction force (N), without the distance. C: $\mu_k$ forgotten. D: $g$ forgotten. E: thinking energy is always conserved; mechanical energy is not conserved with friction.

Question 6medium

A roller-coaster car moves at 5.00 m/s at the top of a 25.0 m high hill. Ignoring friction, what is its speed when it is 10.0 m above the ground?

17.1 m/s
22.1 m/s
14.9 m/s
17.9 m/s
22.7 m/s

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D) 17.9 m/s

$v_f = \sqrt{v_i^2 + 2g(h_i - h_f)} = \sqrt{25.0 + 2(9.80)(15.0)} = 17.9\ \text{m/s}$ . A: the initial speed forgotten. B: the speeds were added directly instead of the energies. C: took the drop as 10.0 m. E: took the drop as 25.0 m.

Question 7easy

A particle moves in one dimension with total mechanical energy E in a potential U(x). At a turning point of its motion:

K = E and U = 0
U = E and K = 0
the force on it is zero
U is at a minimum
K is at a maximum

Show answer

B) U = E and K = 0

At a turning point the particle stops momentarily: $K = 0$ , so $U = E$ . The force is not zero; it is $-dU/dx$ , and it is what turns the particle back. The minimum of $U$ is the stable equilibrium point, where $K$ is a maximum.

Question 8easy

How much energy is equivalent to a mass of 1.00 g? (c = 3.00 × 10⁸ m/s)

9.00 × 10¹⁶ J
3.00 × 10⁵ J
4.50 × 10¹³ J
9.00 × 10⁷ J
9.00 × 10¹³ J

Show answer

E) 9.00 × 10¹³ J

$E = mc^2 = (1.00\times10^{-3}\ \text{kg})(3.00\times10^8)^2 = 9.00\times10^{13}\ \text{J}$ . A: grams were not converted to kilograms. B: $mc$ , no square. C: $\tfrac12 mc^2$ , confusing it with the kinetic energy formula. D: $c$ taken in km/s.

Question 9medium

A 1.50 kg block slides at 4.00 m/s on a frictionless horizontal surface and runs into a spring with k = 240 N/m. What is the maximum compression of the spring (in m)?

Show answer

0.316 m

$\tfrac12 mv^2 = \tfrac12 kx_{max}^2 \Rightarrow x_{max} = v\sqrt{m/k} = 4.00\sqrt{1.50/240} = 0.316\ \text{m}$ . Check: $K = 12.0$ J, $\tfrac12(240)(0.316)^2 = 12.0$ J.

Question 10medium

A 70.0 kg skier starts from rest at the top of a slope 20.0 m high and reaches the bottom at 15.0 m/s. How much mechanical energy (in J) was lost to friction and air drag?

Show answer

5.85 × 10³ J

$E_i = mgh = (70.0)(9.80)(20.0) = 13\,720\ \text{J}$ . $E_f = \tfrac12(70.0)(15.0)^2 = 7875\ \text{J}$ . Loss $= 13\,720 - 7875 = 5845 \approx 5.85\times10^3\ \text{J}$ .

Question 11hard

A small cart rolls without friction from rest down a track and around a vertical loop of radius $R = 0.800\ \text{m}$ . What minimum starting height (above the bottom of the loop) lets it stay on the track at the top of the loop?

1.60 m
2.00 m
2.40 m
0.800 m
1.20 m

Show answer

B) 2.00 m

At the top at least $v^2 = gR$ (normal force zero, only $mg$ is centripetal). Energy: $mgh = mg(2R) + \tfrac12 m(gR) \Rightarrow h = 2.5R = 2.00\ \text{m}$ . A: $2R$ , assuming the speed is zero at the top. C: $3R$ . D: $R$ . E: $1.5R$ .

Verification: All numerical answers and distractors were computed with node (verify/PHY1001/13.mjs), g = 9.80 m/s², c = 3.00 × 10⁸ m/s.

Review card

Mechanical energy: E = K + U = ½ m v² + m g y + ½ k x² If only conservative forces do work: Ki + Ui = Kf + Uf With friction: Kf + Uf = Ki + Ui - f_k d (d = path) ΔE_mech = -f_k d → goes to internal energy (heat) With an external push / motor: Ef = Ei + W_ext Free fall: v = √(2 g h) independent of mass Pendulum: h = L (1 - cos theta) Spring → speed: v = x √(k/m) speed → spring: x = v √(m/k) Ramp + friction: v = √(2 g d (sin - mu cos)) Minimum speed at the top of a loop: v² = g R → h_min = 2.5 R Energy diagram: turning point U = E (K = 0); the region E < U is forbidden E = m c² c = 3.00 × 10⁸ m/s m in kg!