Work, Energy & Power

W = Fs cosθ, where θ is the angle between the force and the displacement.

• Positive (0 ≤ θ < 90°): the force helps the motion. Gravity on a falling ball, the pull on a dragged box.
• Zero (θ = 90°, or no displacement): centripetal force in uniform circular motion, tension in a swinging pendulum string, the normal force on a block sliding on a fixed surface, a porter holding a load still or carrying it horizontally at steady speed (his upward force is perpendicular to the motion).
• Negative (90° < θ ≤ 180°): the force opposes the motion. Friction on a block sliding over a fixed floor, gravity on a ball going up, brakes.

Work is a scalar, measured in joules: 1 J = 1 N m = 10⁷ erg. Other units: 1 kWh = 3.6 × 10⁶ J and 1 eV = 1.6 × 10⁻¹⁹ J.

When F changes with position, split the path into tiny steps dx. On each step the force is nearly constant, so dW = F dx. Adding them gives W = ∫F dx, which is the area under the F–x graph.

• Area above the x-axis counts as positive work; area below counts as negative.
• Break the graph into triangles, rectangles and trapeziums and add the signed areas.
• In vector form, \(W=\int(F_x\,dx+F_y\,dy+F_z\,dz)\). For a constant force this is simply \(F_x\Delta x+F_y\Delta y+F_z\Delta z\).

Work done by gravity depends only on the vertical drop: \(W_g=mgh\) for a fall through h, along any path.

1. For a particle moving along x under a net force F: \(F=ma=m\dfrac{dv}{dt}\).
2. Use the chain rule: \(\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v\dfrac{dv}{dx}\). So \(F=mv\dfrac{dv}{dx}\).
3. Multiply by dx and integrate: \(\displaystyle\int_{x_i}^{x_f}F\,dx=\int_{u}^{v}mv\,dv=\tfrac12mv^2-\tfrac12mu^2\).
4. The left side is the net work. So W_net = ΔK.

Nothing in the derivation assumed F was constant or conservative. The theorem holds for any forces, as long as you include all of them and work in an inertial frame.

Since p = mv, \(K=\dfrac{p^2}{2m}\) and \(p=\sqrt{2mK}\). This link drives most ratio questions:

• Same momentum: K ∝ 1/m. The lighter body has more kinetic energy.
• Same kinetic energy: p ∝ √m. The heavier body has more momentum.
• Percentage changes: if p rises by 20%, K rises by a factor 1.2² = 1.44, i.e. 44%. If K rises by 300% (becomes 4K), p doubles (rises by 100%).
• For small changes only: ΔK/K ≈ 2Δp/p (so a 1% rise in p gives about 2% rise in K).

A resistive force that is constant removes kinetic energy in proportion to distance. If a bullet loses half its speed after penetrating a distance d, it has lost ¾ of its kinetic energy. The remaining ¼ needs only d/3 more. In general, the number of identical planks a bullet can cross is proportional to its initial kinetic energy.

A force is conservative if any one of these equivalent statements holds:

• The work it does between two points does not depend on the path.
• The work it does around any closed path is zero.
• It can be written as \(F_x=-dU/dx\) for some potential-energy function U.

Gravity, the spring force and the electrostatic force are conservative. Friction, air drag and viscous forces are not: their work around a closed loop is always negative, because they oppose the motion everywhere.

Near Earth’s surface U = mgh is valid only for h much smaller than Earth’s radius; for large heights use the gravitation chapter’s −GMm/r.

Use F = −dU/dx: the force is minus the slope. A particle is pushed "downhill" on the U–x graph.

• Equilibrium: F = 0, so dU/dx = 0 (flat points of the curve).
• Stable (point A, a minimum, d²U/dx² > 0): a small displacement brings a restoring force back toward A. The particle oscillates.
• Unstable (point B, a maximum, d²U/dx² < 0): a small displacement gives a force pushing it further away.
• Neutral (region C, flat, d²U/dx² = 0): displaced, it stays put.

With total energy E, the kinetic energy at any x is K = E − U(x). Motion is possible only where E ≥ U. The points where E = U are turning points: the particle stops there and turns back. K (and speed) is largest where U is smallest.

1. To stretch the spring slowly, you apply F_ext = +kx (equal and opposite to the spring force).
2. Work by you from 0 to x: \(\int_0^x kx\,dx=\tfrac12kx^2\). This is the triangle under the F–x line (base x, height kx).
3. From x₁ to x₂: \(\tfrac12k(x_2^2-x_1^2)\), the area of a trapezium.
4. The spring force is the opposite, so its work is \(-\tfrac12k(x_2^2-x_1^2)\). Moving back toward natural length, the spring does positive work.

Since U = ½kx² and F = kx, the stored energy can also be written U = F²/(2k) = ½Fx. These forms are handy when the force, not the extension, is given.

• Smooth floor: ½mv² = ½kx_max², so x_max = v√(m/k).
• Rough floor: friction acts during compression: ½mv² = ½kx² + μmgx. Solve this quadratic for x.
• Dropped from height h onto a vertical spring: the block falls a total h + x, so mg(h + x) = ½kx².
• Two blocks joined by a spring on a smooth floor: at maximum compression they move with the same velocity. Use momentum for that velocity and the kinetic energy difference for ½kx².

• Cutting: k is inversely proportional to the natural length (kL = constant). Cut into n equal parts, each has spring constant nk. Cut in the ratio of lengths L₁ : L₂, the parts have k₁ = k(L/L₁) and k₂ = k(L/L₂).
• Series (end to end): same force in each, extensions add. \(\dfrac{1}{k_s}=\dfrac{1}{k_1}+\dfrac{1}{k_2}\).
• Parallel (side by side, same extension): forces add. \(k_p=k_1+k_2\).
• Energy sharing: in series (same F), U = F²/2k, so the softer spring stores more. In parallel (same x), U = ½kx², so the stiffer spring stores more.

1. Choose the system and the start and end states.
2. Write K + U for each state, using one fixed reference level for potential energy.
3. Add the work of all forces not written as potential energy (friction, applied pushes, air drag): K_i + U_i + W_nc = K_f + U_f.
4. Friction makes W_nc negative; a motor or a person pushing can make it positive.

On any smooth track, the speed after falling through height h is √(2gh) whatever the shape of the track. The time taken does depend on the shape.

A bob on a string of length L is released from rest at angle θ₀ from the vertical. It falls a height L(1 − cosθ₀) to the lowest point.

• Speed at the bottom: \(v=\sqrt{2gL(1-\cos\theta_0)}\).
• Tension at the bottom: T − mg = mv²/L gives \(T=mg(3-2\cos\theta_0)\).
• Released from the horizontal (θ₀ = 90°): v = √(2gL) and T = 3mg.

For a uniform chain of mass M and length L with a fraction 1/n hanging over the edge of a smooth table, the hanging part has mass M/n and its centre of mass is L/(2n) below the table. Pulling it back on the table needs

\[W=\frac{M}{n}\,g\,\frac{L}{2n}=\frac{MgL}{2n^2}.\]

Use the centre of mass of each piece whenever an extended body changes height.

Constant force from rest: a = F/m is constant, v = at, so P = Fv = (F²/m)t. Power grows linearly with time.

Constant power from rest (e.g. an engine at full throttle, no losses):

1. The work done equals the kinetic energy gained: Pt = ½mv², so \(v=\sqrt{2Pt/m}\propto t^{1/2}\).
2. Distance: \(x=\int v\,dt\propto t^{3/2}\).
3. Force F = P/v ∝ t^(−1/2), so the force falls as the body speeds up.

• Lifting water: power = (mass per second) × g × h. If the water also leaves with speed v, add (mass per second) × ½v².
• Vehicle at constant speed on a level road: engine force = resistance f, so P = fv. The top speed with maximum power is v_max = P/f.
• Up an incline: P = (mg sinθ + f)v.
• Efficiency: input power = useful power/η. The difference becomes heat.

Units: 1 W = 1 J/s; 1 horsepower (hp) = 746 W.

Total energy is always conserved; "lost" kinetic energy becomes heat, sound and deformation. Momentum is conserved because the collision forces are internal and act for a very short time. External forces such as gravity or friction are finite, so their impulse during the short impact is negligible (the impulse approximation).

When the bodies stick, the common velocity follows from momentum alone, and the kinetic energy lost is

\[\Delta K=\frac12\,\frac{m_1m_2}{m_1+m_2}\,(u_1-u_2)^2.\]

With the target at rest, the fraction lost is m₂/(m₁ + m₂). Two equal masses lose half their kinetic energy. Not all kinetic energy can be lost: the centre of mass keeps moving, and its kinetic energy (P²/2M) must remain.

Ballistic pendulum: a bullet m embeds in a hanging block M, and the pair rises a height h. Use momentum for the collision, then energy for the swing: \(v=\dfrac{m+M}{m}\sqrt{2gh}\). Do not use energy conservation across the collision itself.

In two dimensions, add the momenta as vectors: the combined body moves along the direction of the total momentum.

An explosion is a collision run backward. A body at rest that splits into two pieces has zero total momentum, so the pieces move in opposite directions with m₁v₁ = m₂v₂. Their kinetic energies share as K₁/K₂ = m₂/m₁: the lighter piece carries more energy. A gun and its bullet are the standard example: the gun recoils slowly but the bullet gets almost all the kinetic energy.

Momentum conservation plus the definition of e give, for a head-on collision,

\[v_1=\frac{(m_1-em_2)u_1+(1+e)m_2u_2}{m_1+m_2},\qquad v_2=\frac{(m_2-em_1)u_2+(1+e)m_1u_1}{m_1+m_2}\]

Putting e = 1 recovers the elastic formulas; e = 0 gives the common velocity. The kinetic energy lost is

\[\Delta K=\frac12\,\frac{m_1m_2}{m_1+m_2}\,(1-e^2)(u_1-u_2)^2.\]

For equal masses with one at rest: v₁ = u(1 − e)/2 and v₂ = u(1 + e)/2.

• Each bounce multiplies the speed by e, so the height is multiplied by e²: h₁ = e²h, h₂ = e⁴h, hₙ = e²ⁿh.
• e can be measured from two heights: e = √(h₁/h).
• Fraction of kinetic energy lost in each bounce = 1 − e².
• Total distance before it stops: h + 2(e²h + e⁴h + …) = \(h\dfrac{1+e^2}{1-e^2}\).
• Total time: with t₀ = √(2h/g), each bounce takes 2eⁿt₀, so T = \(t_0\dfrac{1+e}{1-e}\).

Ball hitting a smooth fixed wall or floor: resolve the velocity into a normal part and a tangential part. The tangential part is unchanged (no friction). The normal part reverses and is multiplied by e. If the ball arrives at angle θ to the normal, it leaves at θ′ with tanθ′ = tanθ/e. For e < 1 the rebound is flatter (θ′ > θ).

Two equal smooth spheres, one at rest, elastic oblique collision: after the collision they move at right angles to each other (θ₁ + θ₂ = 90°). This follows from momentum (a vector triangle) and energy (Pythagoras) together.

1. Let the bob have speed u at the bottom and be at angle θ measured from the lowest point. It has risen r(1 − cosθ), so \(v^2=u^2-2gr(1-\cos\theta)\).
2. Radial equation (toward the centre): T − mg cosθ = mv²/r.
3. Combine: \(T=\dfrac{mu^2}{r}+mg(3\cos\theta-2)\).
4. At the top (θ = 180°): T_top = mu²/r − 5mg. The string stays taut if T_top ≥ 0, so u² ≥ 5gr and v_top² ≥ gr.
5. T_bottom − T_top = (mu²/r + mg) − (mu²/r − 5mg) = 6mg, for any u that completes the circle.

In the limiting case (u = √(5gr)): bottom T = 6mg; side (θ = 90°) v = √(3gr), T = 3mg; top v = √(gr), T = 0. The kinetic energy at the bottom is 5 times that at the top.

• u ≤ √(2gr): the bob does not rise above the centre level. It swings back and forth like a pendulum; the string stays taut.
• √(2gr) < u < √(5gr): the bob rises above the centre, but the tension becomes zero somewhere in the upper half. The string slackens and the bob follows a parabola.
• u ≥ √(5gr): complete circles.

• Light rigid rod (or a bead on a ring): the rod can push, so the bob only needs to reach the top with speed ≥ 0. Minimum bottom speed is √(4gr) = 2√(gr).
• Block released on a smooth loop-the-loop track: it needs v_bottom² ≥ 5gR, so the release height must be at least 5R/2.
• Body sliding off the top of a smooth sphere (from rest at the top): it leaves the surface when cosθ = 2/3, i.e. at a height 2R/3 above the centre (R/3 below the top).

1. Momentum: m₁(u₁ − v₁) = m₂(v₂ − u₂).
2. Kinetic energy: m₁(u₁² − v₁²) = m₂(v₂² − u₂²), i.e. m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂).
3. Divide the second by the first: u₁ + v₁ = v₂ + u₂, so v₂ − v₁ = u₁ − u₂. The relative velocity just reverses (e = 1).
4. Solve the two linear equations to get the boxed results in the formula card above.

Shortcut in practice: write momentum conservation and "separation speed = approach speed". Two linear equations are easier than squaring.

Fraction of kinetic energy transferred to the target: \(\dfrac{4m_1m_2}{(m_1+m_2)^2}\). It is 100% only for equal masses. This is why materials rich in light nuclei (water, heavy water, graphite) are used to slow neutrons in reactors.

Moving wall: a ball hitting a massive wall that approaches at speed V rebounds with speed u + 2V (relative velocity reverses in the wall’s frame).