Energy and Work

Definition. The work done by a constant force $\mathbf{F}$ moving a body through displacement $\mathbf{s}$ is

$W = \mathbf{F} \cdot \mathbf{s} = Fs\cos\theta$

Where $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{s}$.

Derivation. For a force $F$ in the direction of motion:

$W = \int_{s_1}^{s_2} F\,ds$

For constant force: $W = F(s_2 - s_1) = Fs$.

If the force makes angle $\theta$ with the displacement, only the component $F\cos\theta$ in the Direction of motion does work: $W = Fs\cos\theta$. $\blacksquare$

The SI unit of work is the joule (J) = newton-metre (Nm).

Intuition. Work is energy transferred by a force. No work is done if the force is perpendicular To the motion (e.g., the normal reaction does no work on a body sliding on a horizontal surface).

Theorem. The kinetic energy of a body of mass $m$ moving at speed $v$ is

$\mathrm{KE} = \frac{1}{2}mv^2$

Proof. Starting from Newton”s Second Law:

$F = ma = m\frac{dv}{dt} = m\frac{dv}{ds}\frac{ds}{dt} = mv\frac{dv}{ds}$

$F\,ds = mv\,dv$

Integrating from rest ($v=0$) to speed $v$:

$W = \int_0^s F\,ds' = \int_0^v mv'\,dv' = \frac{1}{2}mv^2$

This work equals the kinetic energy gained: $\mathrm{KE} = \tfrac{1}{2}mv^2$. $\blacksquare$

Theorem. The gravitational potential energy of a mass $m$ at height $h$ above a reference level Is

$\mathrm{GPE} = mgh$

Proof. The work done against gravity to raise a mass $m$ through height $h$ is:

$W = F \times h = mg \times h = mgh$

This work is stored as gravitational potential energy. $\blacksquare$

• GPE depends on the choice of reference level ( the ground or lowest point).
• Only changes in GPE are physically meaningful.
• When a body falls through height $h$: loss in GPE $= mgh =$ gain in KE (if no other forces do work).

Roller coasters are a classic application of GPE. A coaster train is hauled to the highest point Using a motor (work done against gravity). From there, GPE converts to KE as it descends, and back To GPE as it climbs the next hill.

Key insight: The maximum speed depends only on the vertical drop, not the track shape (assuming No friction). For a drop of $40\,\mathrm{m}$:

$v_{\max} = \sqrt{2(9.8)(40)} \approx 28\,\mathrm{m/s}$

(approximately $100\,\mathrm{km/h}$). Real coasters never reach this due to friction and air Resistance.

For a loop-the-loop of radius $r$The minimum speed at the top is $\sqrt{gr}$ (from circular Motion). The coaster must enter the loop with enough GPE to reach this speed at the top:

$mgh = \frac{1}{2}m(\sqrt{gr})^2 + mg(2r) \implies h = \frac{5}{2}r$

The entry height must be at least $2.5r$ above the bottom of the loop.

Theorem (Work-Energy Principle). The work done by all forces on a body equals the change in its Kinetic energy:

$W_{\mathrm{net}} = \Delta\mathrm{KE} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$W = \int_{s_1}^{s_2} F\,ds = \int_{s_1}^{s_2} ma\,ds = m\int_{s_1}^{s_2}\frac{dv}{dt}\,ds = m\int_{u}^{v}v'\,dv' = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \quad \blacksquare$

If only conservative forces (gravity) do work:

$\mathrm{KE}_1 + \mathrm{GPE}_1 = \mathrm{KE}_2 + \mathrm{GPE}_2$

When friction is present:

$\mathrm{KE}_1 + \mathrm{GPE}_1 = \mathrm{KE}_2 + \mathrm{GPE}_2 + \mathrm{Work done against friction}$

When multiple bodies interact (e.g., two blocks connected by a string over a pulley), apply the Work-energy principle to the system as a whole:

$W_{\mathrm{external}} = \Delta\mathrm{KE}_{\mathrm{system}} + \Delta\mathrm{GPE}_{\mathrm{system}} + W_{\mathrm{friction}}$

Internal forces (such as tension in a connecting string) do equal and opposite work on the two Masses and cancel out. Only external forces and changes in GPE need be considered.

Example. Two particles of masses $3\,\mathrm{kg}$ and $5\,\mathrm{kg}$ are connected by a light Inextensible string over a smooth pulley. Released from rest, find the speed when the $5\,\mathrm{kg}$ mass has descended $2\,\mathrm{m}$.

Net GPE lost $= (5 - 3)(9.8)(2) = 39.2\,\mathrm{J}$. This equals total KE gained:

$\frac{1}{2}(3 + 5)v^2 = 39.2 \implies 4v^2 = 39.2 \implies v \approx 3.13\,\mathrm{m/s}$

Compare with the force method: $a = \frac{5-3}{5+3}(9.8) = 2.45\,\mathrm{m/s}^2$Then $v = \sqrt{2(2.45)(2)} \approx 3.13\,\mathrm{m/s}$. The energy method avoids solving for tension.

$P = \frac{dW}{dt}$

For a force $F$ on a body moving at speed $v$:

$\boxed{P = Fv}$

Derivation. $P = \dfrac{dW}{dt} = \dfrac{F\,ds}{dt} = Fv$. $\blacksquare$

The SI unit of power is the watt (W) = joule per second (J/s).

$1\,\mathrm{kW} = 1000\,\mathrm{W}$, $1\,\mathrm{MW} = 10^6\,\mathrm{W}$.

At maximum speed (terminal velocity) up a slope, the driving force equals the component of weight Plus friction:

$\frac{P}{v_{\max}} = mg\sin\theta + F_{\mathrm{friction}}$

When a body moves under constant power $P$ (e.g., a car with the throttle fixed), the available Tractive force decreases as speed increases:

$F = \frac{P}{v}$

Since $F = ma$:

$m\frac{dv}{dt} = \frac{P}{v} \implies mv\frac{dv}{dt} = P$

Integrating from rest ($v = 0$) to speed $v$ over time $t$:

$\int_0^v mv'\,dv' = \int_0^t P\,dt' \implies \frac{1}{2}mv^2 = Pt \implies v = \sqrt{\frac{2Pt}{m}}$

Speed increases as $\sqrt{t}$ under constant power, slower than the linear increase under constant Force. This explains why cars feel less responsive at high speeds: the available force at speed $v$ Is only $P/v$.

To find the distance covered:

$s = \int_0^t \sqrt{\frac{2Pt'}{m}}\,dt' = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}t^{3/2} = \frac{2}{3}\sqrt{\frac{2P}{m}}\,t^{3/2}$

The gradient of a kinetic energy-time graph gives the instantaneous power:

$\frac{d(\mathrm{KE})}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt} = Fv = P$

The area under a power-time graph equals the total work done (or total energy transferred).

Average power over a time interval $\Delta t$:

$P_{\mathrm{avg}} = \frac{\Delta W}{\Delta t} = \frac{\Delta\mathrm{KE} + \Delta\mathrm{GPE}}{\Delta t}$

For a body accelerating from rest to speed $v$ in time $t$ under constant force:

$P_{\mathrm{avg}} = \frac{\frac{1}{2}mv^2}{t} = \frac{1}{2}Fv$

This is half the instantaneous power $Fv$ at the end, since velocity increases linearly from $0$ to $v$ while force remains constant. For constant power $P$The average power equals $P$ throughout.

Hooke’s law states that the tension in an elastic spring (or string) is proportional to its Extension from the natural length:

$T = kx$

Where $k$ is the stiffness (spring constant) in $\mathrm{N/m}$And $x$ is the extension.

The force-extension graph is a straight line through the origin. The area under this graph equals The work done stretching the spring.

Theorem. The elastic potential energy stored in a spring of stiffness $k$ extended by $x$ is:

$\mathrm{EPE} = \frac{1}{2}kx^2$

Proof. The force varies from $0$ to $kx$. The work done equals the area under the Force-extension graph (a triangle):

$W = \frac{1}{2} \times x \times kx = \frac{1}{2}kx^2$

By integration:

$\mathrm{EPE} = \int_0^x kx'\,dx' = \left[\frac{1}{2}kx'^2\right]_0^x = \frac{1}{2}kx^2 \quad \blacksquare$

An equivalent form using the tension $T = kx$ at maximum extension is $\mathrm{EPE} = \frac{1}{2}Tx$.

When elastic springs are involved, EPE must be included in the energy balance:

$\mathrm{KE}_1 + \mathrm{GPE}_1 + \mathrm{EPE}_1 = \mathrm{KE}_2 + \mathrm{GPE}_2 + \mathrm{EPE}_2 + W_{\mathrm{friction}}$

A bungee cord behaves like a spring. As a jumper falls, GPE converts to KE until the cord becomes Taut. Once taut, the cord stretches and stores EPE. Energy oscillates between GPE, KE, and EPE until Damping dissipates it.

Example. A jumper of mass $75\,\mathrm{kg}$ leaps from a platform $50\,\mathrm{m}$ above a River. The cord has natural length $25\,\mathrm{m}$ and stiffness $200\,\mathrm{N/m}$. Find the Lowest point reached ($g = 9.8\,\mathrm{m/s}^2$).

At the lowest point, speed $= 0$. If the total distance fallen is $d$ (where $d \gt 25$ since the Cord must be stretched):

$mgd = \frac{1}{2}k(d - 25)^2$

$735d = 100(d - 25)^2 \implies 100d^2 - 5735d + 62500 = 0$

$d = \frac{5735 \pm \sqrt{5735^2 - 4(100)(62500)}}{200} = \frac{5735 \pm 2809}{200}$

$d = 42.72\,\mathrm{m}$ (the root $14.63\,\mathrm{m}$ is rejected since the cord is not yet taut).

The lowest point is $50 - 42.72 = 7.28\,\mathrm{m}$ above the river. The jumper is safe.

A mass $m$ on a spring of stiffness $k$ (smooth surface) forms a simple harmonic oscillator. At the Natural length, all energy is kinetic. At maximum extension $A$ (the amplitude), all energy is EPE:

$E_{\mathrm{total}} = \frac{1}{2}kA^2 = \frac{1}{2}mv_{\max}^2$

This gives $v_{\max} = A\sqrt{k/m}$The maximum speed at the equilibrium position.

• Mechanics
• Energy and Work (Extended)
• Kinematics