Conservation of Energy: Study Guide (Physics 1 3.4)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra and graphs. If you are taking the calculus-based Physics C: Mechanics course, it has its own separate guide.

What energies can a system have?

Start every energy problem by drawing a boundary around the system: the objects you choose to track. Everything outside is the surroundings (or environment). What energy the system can hold depends on what is inside.

System	Energies it can have	Why
One object (a ball on its own)	kinetic only	potential energy needs two interacting objects (Topic 3.3)
Ball + Earth	kinetic and gravitational potential	ball and Earth attract by a conservative force
Block + spring	kinetic and elastic potential	the spring changes shape reversibly
Block + spring + Earth	kinetic, elastic and gravitational potential	two conservative interactions
Sled + snowy slope + Earth	kinetic, gravitational potential and thermal	friction inside the system warms the surfaces

Mechanical energy is the sum of a system’s kinetic and potential energies:

E_mech = K + U

Here K = ½mv² for each moving object, and U is the total of all the potential energy terms (U_s, U_g) in the system.

The energy rule

Energy is conserved in every interaction. It is never created or destroyed. It can only do two things:

1. Change form inside the system. A drop in one type of energy is matched by an equal rise in other types. Falling: U_g goes down, K goes up by the same amount.
2. Cross the boundary. When an outside force does work on the system, energy moves in (positive work) or out (negative work).

Put together:

ΔE_system = W (work done on the system by outside forces)

If W = 0, the system’s total energy is constant. If W is not zero, the system’s energy changes by exactly W. Energy is still conserved overall; it has simply moved between the system and its surroundings.

Choosing the system

The same event can be described in different ways depending on the system you choose. A 0.50 kg ball is dropped from rest and falls 2.0 m. Ignore air resistance. Take U_g = 0 at the landing point.

	System: ball only	System: ball + Earth
Energies inside	K	K and U_g
Is gravity inside or outside?	outside: Earth pulls on the system	inside: an interaction between two system objects
Work by outside forces	W = mgh = 0.50 × 9.8 × 2.0 = +9.8 J	0
ΔK	+9.8 J	+9.8 J
ΔU_g	(not part of this system)	−9.8 J
Does the system’s energy change?	yes, by +9.8 J	no: +9.8 − 9.8 = 0

Both descriptions give the same answer for the ball’s speed. The ball + Earth system is usually easier, because you never calculate gravity’s work: it is already counted as ΔU_g. Choosing a system whose total energy is constant is the main skill of this topic.

When is mechanical energy constant?

The mechanical energy of a system stays constant when both of these are true:

1. No outside force does work on the system (W = 0).
2. No non-conservative interactions happen inside it. Friction, air resistance and crumpling or squashing objects are non-conservative.

Then K_i + U_i = K_f + U_f.

When friction or air resistance acts, some mechanical energy is dissipated: it becomes thermal energy (the surfaces warm up) and sound. It does not come back as kinetic or potential energy. If the surfaces that warm up are inside the system and W = 0, the total energy is still constant:

K_i + U_i = K_f + U_f + ΔE_thermal

Here ΔE_thermal includes any energy carried off as sound. You do not need to calculate how much goes to each.

Energy bar charts

An energy bar chart shows how much of each type of energy the system has at chosen moments. To draw one:

1. Name the system and mark the zero of each potential energy.
2. Pick the moments to compare (start, a middle point, the end).
3. Draw one bar for each energy type at each moment. Bar heights must be to scale, or at least in the right ratio.
4. If W is not zero, add a bar for energy crossing the boundary. If friction acts, add a bar for ΔE_thermal.
5. Check: the total height must be the same at every moment when W = 0.

Worked example 1: speed on a smooth track

Question. A cart is released from rest at the top of a smooth (frictionless) curved track, 1.8 m above the lowest point. Ignore air resistance. (a) Derive an expression for the cart’s speed at the lowest point. (b) Calculate it. (c) Find the speed where the track is 0.80 m above the lowest point.

System: cart + Earth. The track’s normal force is always perpendicular to the motion, so it does no work. No friction. So mechanical energy is constant. Take U_g = 0 at the lowest point.

1. (a) Start: K_i = 0 and U_i = mgh. Lowest point: K_f = ½mv² and U_f = 0.
2. mgh = ½mv². The mass cancels: v = √(2gh).
3. (b) v = √(2 × 9.8 × 1.8) = 5.9 m/s.
4. (c) The cart has dropped 1.8 − 0.80 = 1.0 m. mg(1.8 m) = mg(0.80 m) + ½mv², so v = √(2 × 9.8 × 1.0) = 4.4 m/s.

Interpretation. The answer does not depend on the cart’s mass or on the shape of the track. Only the drop in height matters. Energy cannot tell you how long the trip takes, because that does depend on the shape.

Check. √(2gh) has units √(m/s² × m) = m/s. A bigger drop gives a bigger speed, as it should.

Worked example 2: a spring launcher

Question. A toy launcher fires a 0.050 kg ball straight up. Its spring has k = 400 N/m and is compressed by 0.060 m. The ball leaves the spring when it reaches the spring’s relaxed length. Ignore air resistance and the spring’s mass. Take +y upward and U_g = 0 at the ball’s starting position. (a) How high above its starting position does the ball rise? (b) How fast is it moving as it leaves the spring? (c) Predict the height if the compression is doubled.

System: ball + spring + Earth. No outside work, no friction, so K + U_s + U_g is constant (Figure 1).

1. Energy stored: U_s = ½kΔx² = ½ × 400 × (0.060)² = 0.72 J.
2. (a) At the top, K = 0 and U_s = 0, so all 0.72 J is U_g: mgh = 0.72 J. h = 0.72 ÷ (0.050 × 9.8) = 1.5 m (1.47 m).
3. (b) When the ball leaves the spring it has risen 0.060 m, so U_g = 0.050 × 9.8 × 0.060 = 0.029 J. Then K = 0.72 − 0.029 = 0.69 J. v = √(2K / m) = √(2 × 0.69 ÷ 0.050) = 5.3 m/s.
4. (c) h = kΔx² / (2mg), so h ∝ Δx². Doubling Δx multiplies h by 4: about 5.9 m.

Check. The bar heights in Figure 1 add to 0.72 J at A, B and C. If you forgot U_g at B, you would get 5.4 m/s, slightly too fast.

Worked example 3: a slide with friction

Question. A 40 kg child starts from rest at the top of a playground slide 3.0 m high and reaches the bottom at 6.0 m/s. (a) How much mechanical energy is dissipated? (b) Where does it go? (c) What speed would she reach on a frictionless slide?

System: child + slide + Earth. Take U_g = 0 at the bottom.

1. (a) Start: U_g = mgh = 40 × 9.8 × 3.0 = 1176 J ≈ 1200 J; K = 0.
2. Bottom: K = ½ × 40 × 6.0² = 720 J; U_g = 0.
3. Mechanical energy dissipated = 1176 − 720 = 460 J (456 J), about 39% of the start value.
4. (b) Friction between the child and the slide turns it into thermal energy in her clothes and the slide surface, plus a little sound. Both surfaces are inside the system and no outside force does work, so the total energy of the system is still 1176 J.
5. (c) Without friction, v = √(2gh) = √(2 × 9.8 × 3.0) = 7.7 m/s.

Interpretation. Mechanical energy decreased but energy was not destroyed. The “missing” 456 J is in the warmed surfaces.

Comparing scenarios with energy

Energy methods make some comparisons almost instant.

Three throws from a cliff. Three identical stones are thrown from the same 15 m cliff top at 12 m/s: one straight up, one horizontally, one straight down. Ignore air resistance. With the system stone + Earth, each starts with the same K and the same U_g, and falls the same 15 m. So each lands with the same speed: v = √(12² + 2 × 9.8 × 15) = 21 m/s. Their paths and flight times differ, and the horizontal throw lands at an angle while the other two land moving straight down, but the speed does not depend on any of those.

Testing conservation with data. If mechanical energy is conserved for a low-friction cart with light wheels running down a ramp from rest, then ½mv² = mgh, so v² = 2gh. A graph of v² against h should be a straight line through the origin with slope 2g = 19.6 m/s². A smaller slope is evidence that some mechanical energy is being dissipated. Practice Q7 works through a data set like this.

Common misconceptions

• “A falling ball’s potential energy turns into kinetic energy.” Only for the ball + Earth system. For the ball alone, there is no potential energy: gravity does work from outside.
• “Energy is lost when there is friction.” Mechanical energy decreases; total energy does not. It becomes thermal energy and sound (Worked example 3).
• “Heavier objects reach the bottom of a smooth slope faster.” The mass cancels in mgh = ½mv² (Worked example 1).
• “The steeper or longer path gives a different final speed.” On frictionless paths only the change in height matters.
• “Doubling the spring compression doubles the height.” U_s ∝ Δx², so the height quadruples (Worked example 2).
• “The normal force from a track adds energy.” It acts perpendicular to the motion and does no work.
• “Conservation of energy means a system’s energy can never change.” It changes whenever outside forces do work on it. The choice of system decides this.
• Mixing two zeros. Pick one zero for U_g and use it at every moment in the same problem.

Where this leads

Topic 3.5, Power, asks how fast energy is transferred, not just how much. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also look back at Topic 3.3, Potential Energy, or return to the course roadmap.