Unit 01 · Lesson 06 of 08

Energy has more than one place to hide.

A lifted weight, a speeding cart, and warmer water store energy in different ways. Decide which stores change before simplifying an energy balance.

Account for storage

What you’ll be able to do

Calculate kinetic and gravitational potential energy changes.
Distinguish bulk motion from internal energy.
Recognize energy conversion when mechanical motion stops.

Before you touch the controls

Double an object’s speed, then double its height. Which change multiplies its corresponding energy by four?

Explore the modelChange a control. Follow the response.

Kinetic energy16 J

Potential energy relative to floor58.86 J

Mass2 kgSpeed4 m/sHeight above the reference3 m

The controls specify independent states, not a free trajectory. Internal energy is unchanged; $g = 9.81 \mathrm{m}/\mathrm{s}$ ². The bars share a joule scale and rescale to the larger value.

Bulk motion stores kinetic energy.

For a mass moving at speed v, kinetic energy is $\frac{mV^2}{2}$ . The squared speed is crucial: doubling speed quadruples kinetic energy at fixed mass. Speed must be in m/s when mass is in kg and energy is in J.

KE=\frac12mV^2

Elevation stores gravitational potential energy.

Near Earth’s surface, with approximately constant g, an elevation change stores mgΔz. The zero of elevation is a choice; energy changes are independent of that choice. Raising the same mass twice as far doubles its potential-energy increase.

\Delta PE=mg(z_2-z_1)

Internal energy is a separate store.

Internal energy accounts for microscopic energy within matter, distinct from the bulk motion and elevation of the system. Temperature helps describe it, but temperature alone does not specify total internal energy: mass, material, phase, and state matter. In a suitable constant-specific-heat model, $\Delta U=mc\Delta T$ .

E=U+KE+PE

A worked example

Lifting while accelerating.

A 2 kg object rises 3 m and speeds up from rest to 4 m/s.
$\Delta PE = 2 \times 9.81 \times 3 = 58.86 \mathrm{J}$ ; $\Delta KE = \frac12 \times 2 \times 4$ ² = 16 J.
Its mechanical energy increases by 74.86 J. If internal energy is unchanged, that is also its total energy increase.

Make the idea your own

Predict. Calculate. Question the model.

Try each problem before opening its solution. Then change a value to test whether the reasoning still holds.

01 · Calculate

Double the speed.

A 2 kg cart speeds up from 3 m/s to 6 m/s. Find the kinetic-energy increase, then compare final and initial kinetic energies.

A small hint

Subtract the two squared-speed energies, not the energy evaluated at the speed difference.

Explore the visual solution

Initial kinetic energy9 J

Final kinetic energy36 J

Kinetic-energy increase27 J

For m = 2 kg: ΔKE = ½ × 2 × (6² − 3²).

What if you change it?Final speed6 m/s

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Initial $KE = \frac12 \times 2 \times 3$ ² = 9 J.

Carry this forward. A difference of squares is not the square of a difference.

02 · Calculate

How much power does a lift need?

A lift raises 20 kg at a constant speed through 5 m in 10 s. Neglect losses. Find the energy increase and average input power.

A small hint

At constant speed, kinetic energy does not change.

Explore the visual solution

Gravitational energy increase981 J

Rise = 5 m. Average power over 10 s = 98.1 W. Drawing is schematic.

What if you change it?Vertical rise in 10 s5 m

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

$\Delta PE = 20 \times 9.81 \times 5 = 981 \mathrm{J}$ .

Carry this forward. A force can do work even when acceleration is zero.

03 · Challenge

Where does the falling weight’s energy go?

A 10 kg weight falls 2 m and drives a paddle in 1 kg of water. The weight starts and ends at rest. Assume all lost gravitational energy becomes water internal energy, with no heat loss. Use $c = 4,180 \mathrm{J}$ /(kg·K). Find the water temperature rise.

A small hint

Connect the mechanical-energy loss to mcΔT.

Explore the visual solution

Water temperature increase0.0469 K

Lost gravitational energy = 196.2 J. It enters the water as paddle work.

What if you change it?Weight drop2 m

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

The weight loses $10 \times 9.81 \times 2 = 196.2 \mathrm{J}$ of gravitational energy.

Carry this forward. A temperature increase does not prove that the energy entered as heat.

Your learning path

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