Lesson 4 · 40 min
The Principle of Work and Energy
Add up the work of every force on a particle between two positions, and you know how much its kinetic energy changed. That one scalar equation, \(T_1 + \sum U_{1\to2} = T_2\), finds speeds, stopping distances, spring compressions and unknown forces, without ever finding the acceleration.
Learning objectives
- State the principle of work and energy, \(T_1 + \sum U_{1\to2} = T_2\), with \(T = \tfrac12 mv^2\), and explain where it comes from.
- Apply it with a four-step procedure: positions, free-body diagram, works and kinetic energies, solve.
- Find speeds, distances, spring compressions and constant forces, including problems with several stages.
- Apply it to a system of particles joined by inextensible cords.
The principle
Lesson 1 integrated Newton's second law along the path: \(\int\sum F_t\,ds = \tfrac12 mv_2^2 - \tfrac12 mv_1^2\). Lessons 2 and 3 named the left side: it is the sum of the works of all the forces, \(\sum U_{1\to2}\). Name the right side too. The kinetic energy of a particle of mass \(m\) moving at speed \(v\) is
\[ \colKE{T = \tfrac12 mv^2} \]Then:
Principle of work and energy
\[ \colKE{T_1} + \textstyle\sum U_{1\to2} = \colKE{T_2} \]The particle's kinetic energy at 1, plus the work done by all the forces as it moves from 1 to 2, equals its kinetic energy at 2.
- One scalar equation, so one unknown: a speed, a distance, a spring stiffness, a friction coefficient or a constant force.
- \(T\) is a scalar, never negative, in joules. It depends on the speed only: with components, \(v^2 = v_x^2 + v_y^2\). A particle has the same \(T\) moving up, down or sideways.
- Forces perpendicular to the path do no work, but they still matter: the normal force sets the friction, and Lesson 8 needs it on curved paths.
- Speeds are measured in an inertial frame, which for these problems is the ground.
- Hibbeler and Meriam & Kraige write the work as \(U_{1\text{-}2}\); Beer & Johnston write \(U_{1\to2}\). Same thing.
Procedure for analysis
Four steps
- Positions. Pick position 1 and position 2: where you know the speed and where you want it (or where the particle stops).
- Free-body diagram at a general position between them. Mark each force as doing positive, negative or no work.
- Kinetic energies and works. Write \(T_1\) and \(T_2\), then the work of each force with its sign: \(-W\,\Delta y\), \(-\left(\tfrac12 ks_2^2 - \tfrac12 ks_1^2\right)\), \(-\mu_k N d\), \(F\cos\theta\,\Delta s\).
- Solve and check. Substitute into \(T_1 + \sum U_{1\to2} = T_2\). Check the signs, the units, and whether the answer is reasonable.
Example 4.1 — A crate slides into a spring
A \(20\ \text{kg}\) crate is released from rest on a \(30^\circ\) incline, \(3\ \text{m}\) (measured along the incline) above the free end of a spring with \(k = 2\ \text{kN/m}\). The coefficient of kinetic friction is \(\mu_k = 0.25\). Find (a) the speed of the crate when it reaches the spring and (b) the maximum compression of the spring.
Show solution
Forces. \(W = 196.2\ \text{N}\). Perpendicular to the incline nothing accelerates, so \(N = W\cos 30^\circ = 169.9\ \text{N}\) and \(\colF{F_f} = \mu_k N = 42.48\ \text{N}\), up the incline. The weight's component down the incline is \(W\sin 30^\circ = 98.1\ \text{N}\). \(N\) does no work.
(a) Position 1 at rest, position 2 at the spring, \(3\ \text{m}\) further down:
\[ 0 + \colVg{(98.1)(3)} \colF{{} - (42.48)(3)} = \tfrac12(20)\,v_2^2 \quad\Rightarrow\quad 166.9 = 10\,v_2^2 \quad\Rightarrow\quad v_2 = 4.08\ \text{m/s} \](b) Position 3 at maximum compression \(s\), where the crate is momentarily at rest. The weight and friction act over \(3 + s\), the spring only over \(s\):
\[ 0 + \colVg{98.1(3 + s)} \colF{{} - 42.48(3 + s)} \colVe{{} - \tfrac12(2000)s^2} = 0 \] \[ 1000s^2 - 55.62s - 166.9 = 0 \quad\Rightarrow\quad s = 0.437\ \text{m} \]The other root, \(-0.382\ \text{m}\), has no meaning here. Check: starting instead from position 2, with \(T_2 = 166.9\ \text{J}\), gives the same quadratic, because work adds up over pieces of a path. At \(s = 0.437\ \text{m}\) the spring pushes with \(874\ \text{N}\), far more than \(98.1\ \text{N}\) down the slope plus the largest friction, so the crate bounces back up.
Watch the ledger at maximum compression: \(\colKE{T_2} = 0\), so the three works add up to zero, which is the equation of part (b). After the bounce, the weight's work shrinks as the crate climbs back, the spring's work returns to zero once the crate leaves it, and friction's work only ever grows more negative.
More worked examples
Example 4.2 — Skidding downhill
The \(1200\ \text{kg}\) car of Example 1.1, at \(90\ \text{km/h}\) with \(\mu_k = 0.65\), locks its wheels on a \(6^\circ\) downhill grade instead of a level road. (a) How far does it skid? (b) How far would it skid on the level from twice the speed?
Show solution
(a) Now \(N = W\cos 6^\circ\), and the weight does positive work as the car goes down, \(\colVg{W\sin 6^\circ\,d}\):
\[ \tfrac12 mv_1^2 + \colVg{mg\sin 6^\circ\,d} \colF{{} - \mu_k mg\cos 6^\circ\,d} = 0 \quad\Rightarrow\quad d = \frac{v_1^2}{2g\left(\mu_k\cos 6^\circ - \sin 6^\circ\right)} = \frac{625}{2(9.81)(0.5419)} = 58.8\ \text{m} \]That is \(10\ \text{m}\) more than on the level. The mass still cancels.
(b) On the level, \(d = v_1^2/(2\mu_k g)\) grows with the square of the speed: twice the speed, four times the distance, \(4(49.0) = 196\ \text{m}\).
Example 4.3 — How fast after an angled pull?
The \(20\ \text{kg}\) crate of Example 3.4 starts from rest and is pulled \(6\ \text{m}\) across the floor by \(P = 100\ \text{N}\) at \(30^\circ\) above the horizontal, with \(\mu_k = 0.3\). How fast is it moving then?
Show solution
Example 3.4 found \(\colP{U_P} = 519.6\ \text{J}\) and \(\colF{U_f} = -263.2\ \text{J}\); the weight and \(N\) do no work.
\[ 0 + 519.6 - 263.2 = \tfrac12(20)\,v_2^2 \quad\Rightarrow\quad v_2 = \sqrt{\frac{2(256.5)}{20}} = 5.06\ \text{m/s} \]Systems of particles
For several particles, write the principle for each and add the equations:
Principle of work and energy for a system of particles
\[ \textstyle\sum \colKE{T_1} + \sum U_{1\to2} = \sum \colKE{T_2} \]The sum of the works includes the internal forces between the particles, as well as the external ones.
For bodies joined by an inextensible cord that stays taut, the internal forces do no net work: the cord pulls one body forward exactly as much as it holds the other back, and both ends move the same distance. So the tension drops out. Friction between two bodies of the system that slide on each other does not cancel: it turns some of the energy into heat.
Example 4.4 — Two blocks and a pulley
In Figure 4.2, \(m_A = 10\ \text{kg}\), \(m_B = 5\ \text{kg}\) and the coefficient of kinetic friction under \(A\) is \(0.2\). The system is released from rest. How fast are the blocks moving after \(B\) has fallen \(1.5\ \text{m}\)?
Show solution
The cord is inextensible, so both blocks move \(1.5\ \text{m}\) and have the same speed \(v\). Take the two blocks and the cord as the system: the tension's works cancel. The weight of \(A\) and the normal force on \(A\) are perpendicular to its motion.
\[ 0 + \colVg{m_B g(1.5)} \colF{{} - \mu_k m_A g(1.5)} = \tfrac12(m_A + m_B)\,v^2 \] \[ 73.58 - 29.43 = \tfrac12(15)\,v^2 \quad\Rightarrow\quad v = 2.43\ \text{m/s} \]Treating the blocks separately would have brought in the tension as an extra unknown, with one more equation to solve.
Check your understanding
Key takeaways
- \(T_1 + \sum U_{1\to2} = T_2\), with \(T = \tfrac12 mv^2\): kinetic energy plus the work of every force gives the new kinetic energy.
- Procedure: positions, free-body diagram, works with signs, solve and check. The forces that do no work drop out.
- Distances found this way often lose the mass, and stopping distances grow with \(v^2\).
- For connected bodies, apply the principle to the whole system: an inextensible cord's tension does no net work.
- Next, Lesson 5 asks how fast the work is done.