Formula Sheet · Work and Energy

The work of a force

\[ U_{1\to2} = \int_{s_1}^{s_2} F\cos\theta\,ds = \int_{\rvec_1}^{\rvec_2} \Fvec\cdot d\rvec \]

\(\theta\) is the angle between \(\Fvec\) and the direction of motion. Work is a scalar, in joules: \(1\ \text{J} = 1\ \text{N}\cdot\text{m}\).

The sign of the work
Angle \(\theta\)	Work	The force…
\(0 \le \theta \lt 90^\circ\)	positive	helps the motion
\(\theta = 90^\circ\)	zero	only bends the path
\(90^\circ \lt \theta \le 180^\circ\)	negative	opposes the motion

Constant force, straight path: \(U = F\cos\theta\,\Delta s\). Constant force, any path: \(U = F_x\,\Delta x + F_y\,\Delta y\). A force that varies: the signed area under the \(F\cos\theta\)–\(s\) graph.

No work: a force perpendicular to the motion (normal force of a fixed surface, string of a pendulum), or one whose point of application does not move (static friction on a rolling wheel, a fixed pin).

Work of weight, springs and friction

The work of common forces between positions 1 and 2
Force	\(U_{1\to2}\)
\(\colVg{\text{Weight}}\) (\(y\) up)	\(\colVg{-W(y_2 - y_1) = -W\,\Delta y}\)
\(\colVe{\text{Linear spring}}\)	\(\colVe{-\left(\tfrac12 ks_2^2 - \tfrac12 ks_1^2\right)}\)
\(\colF{\text{Kinetic friction}}\)	\(\colF{-\mu_k N\,d}\) (\(N\) constant, \(d\) = distance slid)
\(\colP{\text{Constant pull}}\) \(P\) at \(\alpha\)	\(\colP{P\cos\alpha\,d}\)

Weight: only the change in height matters, not the path.
Spring: \(s\) is measured from the unstretched length, stretch or compression alike. \(\tfrac12 k(s_2^2 - s_1^2) \ne \tfrac12 k(s_2 - s_1)^2\).
Friction: find \(N\) from the free-body diagram; it is \(mg\) only on a level surface with no other vertical force. On an incline \(N = mg\cos\theta\).

The principle of work and energy

Kinetic energy \(\colKE{T = \tfrac12 mv^2}\) (scalar, never negative)

\[ \colKE{T_1} + \textstyle\sum U_{1\to2} = \colKE{T_2} \]

Procedure: (1) choose positions 1 and 2; (2) free-body diagram at a general position; (3) the work of every force that does work, with its sign; (4) \(T_1\) and \(T_2\); (5) solve for the one unknown.

Finds speeds, distances and spring deformations directly; not times or accelerations.
Several stages: add the work stage by stage, or apply the principle once from start to finish.
Connected bodies: write it for the system. Inextensible cords and smooth pins do no net work.

Power and efficiency

Power

\[ \colP{P} = \frac{dU}{dt} = \Fvec\cdot\vvec = Fv\cos\theta \]

Efficiency

\[ \varepsilon = \frac{P_\text{out}}{P_\text{in}} \lt 1 \]

Use the force at that instant: an accelerating load needs \(F = m(g + a)\), not \(mg\).
Machines in series: \(\varepsilon = \varepsilon_1\varepsilon_2\varepsilon_3\cdots\)
Constant power from rest, no losses: \(\tfrac12 mv^2 = Pt\).
\(1\ \text{W} = 1\ \text{J/s}\); \(1\ \text{hp} = 745.7\ \text{W}\); \(1\ \text{kW}\cdot\text{h} = 3.6\ \text{MJ}\) (energy).

Conservative forces and potential energy

A force is conservative when its work depends only on the end points, so its work around any closed path is zero. Weight and spring forces are; friction, drag and pushes are not.

Gravitational

\[ \colVg{V_g = Wy} \]

Elastic

\[ \colVe{V_e = \tfrac12 ks^2} \]

\(y\) is measured upward from a datum you choose (\(V_g \lt 0\) below it); \(V_e \ge 0\) always. Far from Earth: \(V_g = -GM_em/r\).

\[ U_{1\to2} = V_1 - V_2, \] \[ F = -\frac{dV}{ds} \]

Conservation of energy

\[ \colKE{T_1} + \colVg{V_{g1}} + \colVe{V_{e1}} + \textstyle\sum \colF{U_{nc}} = \colKE{T_2} + \colVg{V_{g2}} + \colVe{V_{e2}} \]

\(\sum U_{nc}\) is the work of the nonconservative forces (friction, drag, pushes); when it is zero, \(T + V\) is conserved. Friction makes it negative: mechanical energy lost as heat.

Count each conservative force once: as a potential energy or as a work, never both.
Pendulum (from rest at \(\theta_0\)): \(v^2 = 2gL(\cos\theta - \cos\theta_0)\).
Spring launcher straight up: \(\tfrac12 ks^2 = mgh\), with \(h\) measured from the release position.

Energy on curved paths

Two equations: energy gives \(v^2\) at the point; then Newton's law along the normal, toward the center of curvature, gives the force.

\[ \textstyle\sum F_n = m\dfrac{v^2}{\rho} \]

Standard results on curved paths
Situation	Result
Loop, top: just makes it	\(N = 0\), \(v^2 = gR\), \(h_{\min} = \tfrac52 R\)
Loop, bottom (from \(h\))	\(N = mg\left(1 + 2h/R\right)\); \(6mg\) from \(\tfrac52 R\)
Pendulum string tension	\(T_s = mg\left(3\cos\theta - 2\cos\theta_0\right)\)
Crest of radius \(\rho\)	\(N = m(g - v^2/\rho)\); leaves if \(v^2 \gt g\rho\)
Leaving a smooth dome	\(\cos\theta = \tfrac23 + v_0^2/(3gR)\); \(48.19^\circ\) from rest

A surface or string can only push (or pull): when the \(N\) you compute has the wrong sign, the particle has left the path. A rod or a bead on a wire can do both.

Choosing a method

When to use energy methods and when to use Newton's second law
Use	When the question asks for
Work and energy	a speed at a position, a distance, a height, a spring compression; forces that vary with position
Conservation of energy	the same, when only weights and springs do work (or the nonconservative work is known)
\(\sum \Fvec = m\avec\)	an acceleration, a time, a force at an instant (often with energy for \(v^2\))
Power	how fast work is done: motors, engines, efficiencies

Common mistakes

Dropping a sign. Friction's work is always negative; the weight's is negative going up.
\(N = mg\) everywhere. Not on an incline, not with an angled push or pull, not on a curve.
\(-\tfrac12 k(s_2 - s_1)^2\) for a spring that starts deformed. Square each deformation first.
Measuring \(s\) from the wrong place. Deformation is from the unstretched length, not from the starting position.
Double counting: \(V_g\) and \(U_W\) for the same weight.
A speed in km/h. Divide by 3.6 before squaring.
\(P = mgv\) for an accelerating load. Use the actual force, \(m(g + a)\).
Assuming a body stays on its path without checking that \(N \ge 0\).

A solution, step by step

A \(20\ \text{kg}\) crate is released from rest on a \(30^\circ\) incline (\(\mu_k = 0.25\)), \(3\ \text{m}\) above a spring with \(k = 2\ \text{kN/m}\). Find the maximum compression \(s\).

Positions: 1 at release (\(T_1 = 0\)); 2 at maximum compression (\(T_2 = 0\)).
Free-body diagram: \(W\), \(N = mg\cos 30^\circ = 169.9\ \text{N}\), friction \(\mu_k N = 42.48\ \text{N}\) up the slope, spring force once in contact.
Works over \(3 + s\): \(\colVg{U_W = 98.1(3 + s)}\), \(\colF{U_f = -42.48(3 + s)}\); over \(s\) only: \(\colVe{U_s = -1000s^2}\).
Principle: \(0 + 55.62(3 + s) - 1000s^2 = 0\).
Solve and check: \(s = 0.437\ \text{m}\) (the other root is negative). Units: J throughout.

The full solution is Example 4.1 in Lesson 4

Check values

Test your method, calculator or code on these before trusting it with a new problem (\(g = 9.81\ \text{m/s}^2\)).

Worked results from the lessons
Skid: \(90\ \text{km/h}\), \(\mu_k = 0.65\), level	\(d = 49.01\ \text{m}\)
\(20\ \text{kg}\) crate, \(30^\circ\), \(\mu_k = 0.25\), \(3\ \text{m}\) to a \(2\ \text{kN/m}\) spring	\(v = 4.085\ \text{m/s}\); \(s_{\max} = 0.437\ \text{m}\)
Hoist: \(300\ \text{kg}\) at \(2\ \text{m/s}\), \(\varepsilon = 0.8\)	\(P_\text{in} = 7.36\ \text{kW}\)
Pendulum released from horizontal	bottom: \(v = \sqrt{2gL}\), \(T_s = 3mg\)
\(k = 400\ \text{N/m}\), \(s = 0.06\ \text{m}\), \(50\ \text{g}\) ball up	\(h = 1.468\ \text{m}\)

Units

\(1\ \text{J} = 1\ \text{N}\cdot\text{m} = 1\ \text{kg}\cdot\text{m}^2/\text{s}^2\). \(36\ \text{km/h} = 10\ \text{m/s}\). \(\text{N/m} \times \text{m}^2 = \text{J}\).

Worked in Lesson 4, Lesson 5 and Lesson 7

Symbols and units

The symbols used in the module and their SI units
Symbol	Meaning	Unit
\(\colKE{T}\)	kinetic energy \(\tfrac12 mv^2\)	J
\(U_{1\to2}\)	work done from position 1 to 2	J
\(\colVg{V_g},\ \colVe{V_e}\)	gravitational, elastic potential energy	J
\(U_{nc}\)	work of nonconservative forces	J
\(\colP{P}\), \(\varepsilon\)	power; efficiency	W; none
\(k\), \(s\)	spring stiffness; deformation	N/m; m
\(\mu_k\), \(N\)	kinetic friction coefficient; normal force	none; N
\(\rho\), \(\kappa = 1/\rho\)	radius of curvature; curvature	m; 1/m