Here are questions that often come up while learning this chapter — with short, clear answers. Questions are ordered from foundational to advanced. If you have a question not listed, reach us via the contact page.
Work: the basics
1. Why is work the dot product \( \mathbf{F} \cdot \mathbf{d} \), not just the force times distance?
Because only the component of force in the direction of motion does work. If a force is perpendicular to motion (e.g., normal force on a sliding box), it does zero work — no energy is transferred. The dot product automatically picks out the parallel component: \( W = Fd\cos\theta \) where \( \theta \) is the angle between force and displacement. This is why work is a scalar (number), not a vector.
2. Why is work negative when friction opposes motion?
Negative work means energy is being taken out of the system. If friction opposes motion, friction removes kinetic energy from the object (converting it to heat). So \( W_{\text{friction}} = -f \cdot d < 0 \). The negative sign is the mathematical way to say "this force reduces the object's energy."
3. Can the work done by one force be positive while another's is negative?
Yes, absolutely. Gravity might do positive work (pulling an object downward), while friction does negative work (opposing the motion). The net work is the sum: \( W_{\text{net}} = W_{\text{gravity}} + W_{\text{friction}} \). This net work equals the change in kinetic energy.
4. Is work the same thing as effort?
No. Effort is muscular (subjective). Work is physics (objective). Example: if you hold a heavy box stationary, you're expending effort (your muscles are working), but physics work is zero — no displacement, so \( W = F \cdot 0 = 0 \). The box's kinetic energy doesn't change.
Kinetic energy
5. Why is kinetic energy \( \frac{1}{2}mv^2 \) and not something else, like \( mv \)?
The factor \( \frac{1}{2} \) and the \( v^2 \) (not \( v \)) come from integration of the work–energy theorem. If you integrate Newton's second law \( F = ma = m(dv/dt) \) over distance, you get:
\[ W = \int F \, dx = \int m\frac{dv}{dt} \, dx = \int mv \, dv = \frac{1}{2}m(v_f^2 - v_i^2) \]
The \( \frac{1}{2} \) and squaring come naturally from calculus. Kinetic energy is \( \frac{1}{2}mv^2 \) because that's what the math demands when you define energy as "what the work theorem measures."
6. Why does kinetic energy depend on \( v^2 \) and not just \( v \)?
Because doubling your speed quadruples the stopping distance (for fixed friction). A car going 100 km/h takes 4 times as long to stop as a car going 50 km/h. The work–energy theorem says \( W_{\text{friction}} = \Delta KE \). If stopping distance scales as \( v^2 \), then kinetic energy must scale as \( v^2 \) too. Hence \( KE = \frac{1}{2}mv^2 \).
7. Can kinetic energy be negative?
No. Kinetic energy is \( KE = \frac{1}{2}mv^2 \), and \( m > 0 \) and \( v^2 \geq 0 \) always. An object at rest (\( v = 0 \)) has zero kinetic energy, not negative. Kinetic energy is always non-negative.
8. What does the work–energy theorem really say in simple terms?
Net work = change in kinetic energy. All forces combined do work equal to the change in the object's kinetic energy. If work is positive, the object speeds up. If negative, it slows down. If zero, speed stays the same.
Potential energy
9. What's the difference between gravitational potential energy \( mgh \) and the general formula \( -\frac{GMm}{r} \)?
- \( mgh \) is a linear approximation near Earth's surface. It's valid when \( h \ll R_{\text{Earth}} \) (height is small compared to Earth's radius). It's simple and accurate for everyday problems (falling ball, jumping).
- \( -\frac{GMm}{r} \) is the exact formula that works at any distance. It accounts for the fact that gravity weakens as \( 1/r^2 \) at large distances. It shows that PE goes to zero at infinity (defining the reference point).
For example, at Earth's surface (\( r = R_E \)): \( PE = -\frac{GMm}{R_E} \). But if you move up by a small height \( h \), the change is:
\[ \Delta PE = mgh \approx -\frac{GMm}{R_E + h} + \frac{GMm}{R_E} \]
For small \( h \), this simplifies to \( mgh \). So \( mgh \) is just the first-order term of the exact formula.
10. Why is gravitational PE defined as \( -\frac{GMm}{r} \), not positive?
Convention. We choose the zero of potential energy at infinity (\( r = \infty \)). Any finite distance \( r \) gives \( PE < 0 \). This is a bound state — negative energy means the object is trapped (bound) by gravity. If you want to escape Earth's gravity completely, you need to add energy to make PE = 0. This sign convention is natural for gravity but can feel weird at first.
11. Can potential energy ever be negative?
Yes. Gravitational PE is negative everywhere near a planet (by our convention). A bound object has negative PE. Elastic PE (springs) is always non-negative because \( PE_{\text{spring}} = \frac{1}{2}kx^2 \geq 0 \). The sign of PE depends on how you choose your reference point (zero). Total mechanical energy can be negative too — it just means the object is trapped.
Conservative forces and energy conservation
12. What's a conservative force in one sentence?
A force is conservative if the work it does is independent of the path — only start and end points matter. Gravity and springs are conservative. Friction is not (work depends on path length).
13. Why is friction not conservative?
Because the work done by friction depends on how long the path is. Drag an object from A to B via a straight line: friction does \( W_1 = -f \cdot d_{\text{straight}} \). Drag via a long, zigzag path: \( W_2 = -f \cdot d_{\text{zigzag}} \), with \( d_{\text{zigzag}} > d_{\text{straight}} \). So \( W_1 \neq W_2 \) — friction's work depends on the path. This is why friction cannot be written as "change in potential energy."
14. Does mechanical energy have to be conserved in real situations?
No. Mechanical energy (KE + PE) is only conserved when no non-conservative forces act (no friction, air resistance, or external work). In reality:
- Friction converts mechanical energy → heat
- Air resistance → heat and turbulence
- Inelastic collisions → permanent deformation (energy "lost")
So use energy conservation carefully. Check: "Are there any dissipative forces?" If yes, energy isn't conserved — it's converted to heat/sound/deformation.
15. In an inelastic collision, where does the "lost" energy go?
Into permanent deformation, heat, and sound. Two cars collide and crumple. Kinetic energy (before collision) is large. After collision, the crumpled cars move slower, so kinetic energy is smaller. The "missing" energy went into:
- Bending metal (work against material forces)
- Heat from friction inside the deforming material
- Sound (vibrations radiating away)
Mechanical energy is not conserved in inelastic collisions. But total energy (including heat and internal energy) is conserved—that's the first law of thermodynamics.
Power
16. What's the difference between power and energy?
- Energy is a quantity (joule, J). It measures how much work you can do or how much has been done.
- Power is a rate (watt, W = J/s). It measures how fast energy is transferred.
Analogy: water and tap. Energy is the volume of water (liters). Power is the flow rate (liters/second). A large tank (high energy) flowing through a narrow opening (low power) takes a long time to empty.
17. Can power be negative?
Yes. Negative power means energy is being removed from the system. When a car brakes, the friction force does negative work, so power \( P = \mathbf{F} \cdot \mathbf{v} < 0 \) (force opposes velocity). The car is losing kinetic energy.
18. Why is power \( P = \mathbf{F} \cdot \mathbf{v} \) and not \( P = \mathbf{F} \cdot \mathbf{d} \)?
Because power is a rate (per unit time). Work is \( W = \mathbf{F} \cdot \mathbf{d} \). Power is work per unit time:
\[ P = \frac{W}{\Delta t} = \frac{\mathbf{F} \cdot \mathbf{d}}{\Delta t} = \mathbf{F} \cdot \frac{\mathbf{d}}{\Delta t} = \mathbf{F} \cdot \mathbf{v} \]
Velocity is displacement per time, so \( \mathbf{v} = \mathbf{d}/\Delta t \). That's why power has velocity in it.
19. A 100 W motor and a 200 W motor do the same job. Why would you ever choose the 100 W motor?
If the job takes the same time, the 200 W motor does it faster. But if you have plenty of time, the 100 W motor is fine and uses less electricity. Or if your power source is limited (e.g., solar panel on a camping trip), lower power might be your only option. Also, lower power means less heat dissipation, so the 100 W motor is cheaper to cool.
Real-world applications
20. In a hydroelectric dam, how does the height of water relate to electricity generated?
The water at the top of the dam has gravitational PE: \( PE = mgh \). When it falls and spins the turbine, PE converts to KE, then to rotational energy in the turbine, then to electrical energy via a generator. Higher dam = more PE = more energy available = more electricity. A dam twice as high can generate roughly twice as much energy (ignoring losses). This is why dams are built on high mountains — height directly translates to energy available.
Question not listed here?
- Reach us via the contact page or via comments on any section.
- We add frequently-asked questions to this page.
Previous chapter: 7.6 Power and Energy Transfer
Next chapter: Chapter 8 — Momentum and Collisions (coming soon)
Have a question? 🤔
If something isn't clear or you have a question, ask it here. The answer will be published on this page.