Lesson 1 · 20 min
Why Energy Methods?
A roller coaster crests a hill, a crate slides into a bumper, a car skids to a stop. The question is usually how fast something is moving at some point, or how far it goes before it stops. Newton's second law can answer it, but only after an integration along the path. Do that integration once, in general, and you get the principle of work and energy: one scalar equation that connects speed and position directly.
Learning objectives
- Explain how integrating \(\sum F_t = m a_t\) along the path relates the speeds at two positions without involving time.
- Recognize the questions that energy methods answer directly, and the ones that still need \(\sum\Fvec = m\avec\).
- Read the module's color code: one color for each kind of energy, shared by the force that exchanges it.
Two slides, one speed
Two frictionless slides start at the same height \(h = 3\ \text{m}\) and end on the same level floor. Slide A is a straight ramp. Slide B drops steeply at first and then flattens out. Identical small blocks are released from rest at the top of each. Before you run the race, make a prediction.
Two things happen in Figure 1.1. The speeds at the bottom are equal, whatever the shape of the slide. The times are not: slide B wins by about \(0.09\ \text{s}\). The first result is what energy methods are built to find. The second is something they cannot see, because time never enters them.
Newton's second law, integrated along the path
Along the path of a particle, the tangential component of Newton's second law is \(\sum F_t = m a_t\). The chain rule writes the tangential acceleration in terms of position instead of time:
\[ a_t = \frac{dv}{dt} = \frac{dv}{ds}\,\frac{ds}{dt} = v\,\frac{dv}{ds} \]So \(\sum F_t\,ds = m v\,dv\). Integrate both sides from position 1 to position 2:
Newton's second law, integrated along the path
\[ \int_{s_1}^{s_2} \textstyle\sum F_t\,ds = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 \]Speeds at two positions, connected by the forces along the path between them. No time, no acceleration.
The left side is the work done by the forces, the subject of Lessons 2 and 3. The right side is the change in the kinetic energy \(\colKE{T = \tfrac12 mv^2}\). Lesson 4 turns this into the working form \(T_1 + \sum U_{1\to2} = T_2\).
Only the tangential component \(F_t\) appears. A force perpendicular to the path, such as the push of a smooth slide, drops out. On either slide the only tangential force is the component of the weight along the slope, and for any shape its integral is \(\colVg{W h}\). That is why the two speeds in Figure 1.1 are equal.
Example 1.1 — A skid distance, two ways
A \(1200\ \text{kg}\) car traveling at \(90\ \text{km/h}\) locks its wheels on a level road. The coefficient of kinetic friction between the tires and the road is \(\mu_k = 0.65\). How far does it skid?
Show solution
First, \(v_1 = 90/3.6 = 25\ \text{m/s}\). The road pushes up with \(N = mg\), so friction is \(\colF{F_f} = \mu_k m g\), against the motion.
With Newton's second law and kinematics. \(\sum F_x = m a\) gives \(-\mu_k m g = m a\), so \(a = -\mu_k g = -6.377\ \text{m/s}^2\), constant. Then \(v^2 = v_1^2 + 2a\,d\) with \(v = 0\):
\[ d = \frac{v_1^2}{2\mu_k g} = \frac{25^2}{2(0.65)(9.81)} = 49.0\ \text{m} \]With the integrated law. The only force along the path is friction, \(F_t = -\mu_k m g\), so the left side is \(-\mu_k m g\,d\):
\[ -\mu_k m g\,d = 0 - \tfrac12 m v_1^2 \quad\Rightarrow\quad d = \frac{v_1^2}{2\mu_k g} = 49.0\ \text{m} \]Same answer. The second route never needed the acceleration or a constant-acceleration formula, and it still works when the force varies along the path, where \(v^2 = v_1^2 + 2ad\) does not apply. Notice too that the mass cancels: a heavier car with the same tires skids just as far.
What energy methods answer, and what they don't
Energy methods are the tool of choice whenever the question links a speed to a position. They are not the only tool, and the integrated law above shows why: time has been integrated away.
| You want | Use | Where |
|---|---|---|
| A speed at a position, or the distance to stop | Work and energy | Lessons 2–4, 6–7 |
| A normal force or cord tension on a curved path | Energy for \(v\), then \(\sum F_n = m v^2/\rho\) | Lesson 8 |
| How fast a motor or engine must do work | Power, \(P = \Fvec\cdot\vvec\) | Lesson 5 |
| An acceleration, or a force at one instant | Newton's second law | Earlier in the course |
| A time, or forces that change with time | Newton's second law with kinematics, or impulse and momentum | Next topic |
Color tells you where the energy goes
Every figure, readout and highlighted formula in this module uses the same colors. Each kind of energy has a color, and the force that exchanges energy with it is drawn in that color. When you see an orange arrow, friction is taking energy out of the motion.
- Kinetic energy and velocity\(\colKE{T = \tfrac12 mv^2}\) and the velocity arrow \(\colKE{\vvec}\)
- GravityThe weight \(\colVg{W}\), its work \(\colVg{U_W}\) and the potential energy \(\colVg{V_g}\)
- SpringsThe spring force \(\colVe{F_s}\), its work \(\colVe{U_s}\) and the potential energy \(\colVe{V_e}\)
- FrictionThe friction force \(\colF{F_f}\) and the energy it dissipates as heat
- Applied forces and powerA push or pull \(\colP{P}\), its work and the power it delivers
- Forces that do no workThe normal force \(N\) of a fixed surface, drawn in black
The road ahead
- Lesson 2 defines the work of a force and reads it as an area. Lesson 3 works out the forces you will meet most: the weight, springs and friction.
- Lesson 4 states the principle of work and energy, \(T_1 + \sum U_{1\to2} = T_2\), with a procedure, and extends it to connected bodies.
- Lesson 5 asks how fast work is done: power and efficiency.
- Lessons 6 and 7 store the work of weights and springs as potential energy and conserve the total.
- Lesson 8 combines energy with \(\sum F_n = m v^2/\rho\) for loops, pendulums and smooth domes.
Check your understanding
Key takeaways
- With \(a_t = v\,dv/ds\), Newton's second law integrated along the path gives \(\int\sum F_t\,ds = \tfrac12 mv_2^2 - \tfrac12 mv_1^2\): speed and position, with no time.
- Only tangential components of force contribute. Forces perpendicular to the path, like the push of a smooth slide, drop out.
- Energy methods answer "how fast here?" and "how far?". They cannot answer "how long?".
- Next, Lesson 2 defines the left side of the integrated law: the work of a force.