Energy and Work (Extended)

This document provides a thorough treatment of the work-energy principle, power, conservation of Energy, and elastic potential energy with proofs and applications.

The work done by a constant force $\mathbf{F}$ acting on a body that undergoes a displacement $\mathbf{s}$ is:

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

Where $\theta$ is the angle between the force and the displacement.

The SI unit of work is the joule ($\mathrm{J} = \mathrm{Nm}$). Work is a scalar quantity.

Special cases:

• $\theta = 0^\circ$: $W = Fs$ (force in the direction of motion).
• $\theta = 90^\circ$: $W = 0$ (force perpendicular to motion — no work done).
• $\theta = 180^\circ$: $W = -Fs$ (force opposing motion).

For a variable force in one dimension:

$W = \int_{x_1}^{x_2} F\,dx$

Proof. For a small displacement $\delta x$The work done is approximately $F\,\delta x$. In the Limit as $\delta x \to 0$:

$W = \lim_{\delta x \to 0} \sum F\,\delta x = \int_{x_1}^{x_2} F\,dx \quad \blacksquare$

Lifting a mass $m$ through a vertical height $h$:

$W = mgh$

This is independent of the path taken (gravitational force is conservative).

Proof. Consider a general path from height $h_1$ to height $h_2$. The gravitational force is $-mg\hat{\mathbf{j}}$. The work done by gravity is:

$W = \int \mathbf{F} \cdot d\mathbf{s} = \int_{h_1}^{h_2} (-mg)\,dh = -mg(h_2 - h_1) = mgh_1 - mgh_2$

The work done against gravity is $mg(h_2 - h_1) = mgh$. $\blacksquare$

Friction is a non-conservative force. The work done by friction depends on the path:

$W_{\mathrm{friction}} = -\mu R \times d$

Where $d$ is the total distance travelled along the surface (not the displacement). Friction always Does negative work (it opposes motion), so it always removes energy from the system.

The kinetic energy of a body of mass $m$ moving with speed $v$ is:

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

Derivation from Newton’s second law. Starting from $F = ma$ and using $a = \dfrac{dv}{dt}$:

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

$F\,ds = mv\,dv$

Integrating:

$\int_{s_1}^{s_2} F\,ds = \int_{v_1}^{v_2} mv\,dv = \left[\frac{1}{2}mv^2\right]_{v_1}^{v_2}$

The left-hand side is the work done by the force, so:

$W = \Delta\mathrm{KE} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

This is the work-energy theorem. $\blacksquare$

Problem. A car of mass $1200\;\mathrm{kg}$ accelerates from $15\;\mathrm{m\,s^{-1}}$ to $25\;\mathrm{m\,s^{-1}}$ over a distance of $200\;\mathrm{m}$ on a level road. Find the average Driving force, given that the total resistance to motion is $400\;\mathrm{N}$.

Work-energy theorem:

$(F - 400) \times 200 = \frac{1}{2}(1200)(25^2 - 15^2)$

$(F - 400) \times 200 = 600(625 - 225) = 600 \times 400 = 240000$

$F - 400 = 1200 \implies F = 1600\;\mathrm{N}$

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

$\boxed{\mathrm{GPE} = mgh}$

This is valid near the Earth’s surface where $g$ is approximately constant.

The choice of reference level (where $\mathrm{GPE} = 0$) is arbitrary. Only changes in GPE Have physical significance:

$\Delta\mathrm{GPE} = mg\Delta h$

The work-energy principle states that the work done by the resultant force on a body equals the Change in its kinetic energy:

$W_{\mathrm{net}} = \Delta\mathrm{KE}$

When conservative forces (gravity, elastic forces) are present:

$W_{\mathrm{non-conservative}} = \Delta\mathrm{KE} + \Delta\mathrm{GPE} + \Delta\mathrm{EPE}$

Or equivalently:

$\Delta\mathrm{KE} + \Delta\mathrm{GPE} + \Delta\mathrm{EPE} = W_{\mathrm{external}}$

Problem. A particle of mass $5\;\mathrm{kg}$ is projected up a rough inclined plane at $30^\circ$ to the horizontal with speed $8\;\mathrm{m\,s^{-1}}$. The coefficient of friction is $0.3$. Find how far up the plane the particle travels before coming to rest.

$R = 5g\cos 30^\circ$

$F = \mu R = 0.3 \times 5g\cos 30^\circ = 1.5g\cos 30^\circ \approx 12.74\;\mathrm{N}$

Work-energy: loss of KE $=$ work done against gravity $+$ work done against friction.

$\frac{1}{2}(5)(64) = 5g \times d\sin 30^\circ + 12.74d$

$160 = 24.5d + 12.74d = 37.24d$

$d = \frac{160}{37.24} \approx 4.30\;\mathrm{m}$

Principle. Energy cannot be created or destroyed, only transformed from one form to another.

For a mechanical system with no friction or other dissipative forces:

$\mathrm{KE} + \mathrm{GPE} + \mathrm{EPE} = \mathrm{constant}$

Problem. A simple pendulum has a bob of mass $0.5\;\mathrm{kg}$ on a string of length $1.5\;\mathrm{m}$. It is released from rest when the string makes an angle of $40^\circ$ with the vertical. Find the Speed of the bob at the lowest point, neglecting air resistance.

Height gain: $h = 1.5 - 1.5\cos 40^\circ = 1.5(1 - \cos 40^\circ) \approx 1.5(1 - 0.766) = 0.351\;\mathrm{m}$.

Conservation of energy: $\mathrm{GPE}_{\mathrm{top}} = \mathrm{KE}_{\mathrm{bottom}}$.

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.351} = \sqrt{6.88} \approx 2.62\;\mathrm{m\,s^{-1}}$

Problem. A roller coaster car of mass $500\;\mathrm{kg}$ starts from rest at point $A$, $30\;\mathrm{m}$ Above the ground. It descends to point $B$ at ground level, then rises to point $C$ at $20\;\mathrm{m}$ Above ground. The average frictional force is $200\;\mathrm{N}$ and the total track length from $A$ To $C$ is $300\;\mathrm{m}$. Find the speed at $C$.

$\mathrm{GPE}_A + \mathrm{KE}_A = \mathrm{GPE}_C + \mathrm{KE}_C + W_{\mathrm{friction}}$

$500g(30) + 0 = 500g(20) + \frac{1}{2}(500)v^2 + 200 \times 300$

$147000 = 98000 + 250v^2 + 60000$

$250v^2 = 147000 - 158000 = -11000$

Since $250v^2$ cannot be negative, the car cannot reach point $C$. It comes to rest before Reaching $C$.

Let us find how far along the track it travels before stopping (let this be $d$ metres from $A$ At height $h$):

$500g(30) = 500gh + 200d$

Without more information about the track profile, we cannot determine the exact stopping point. This illustrates the importance of knowing the track geometry.

Power is the rate of doing work:

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

For a constant force $F$ moving a body at velocity $v$:

$\boxed{P = Fv}$

The SI unit of power is the watt ($\mathrm{W} = \mathrm{J\,s^{-1}}$).

$P = \frac{dW}{dt} = \frac{d}{dt}(\mathbf{F} \cdot \mathbf{s}) = \mathbf{F} \cdot \frac{d\mathbf{s}}{dt} = \mathbf{F} \cdot \mathbf{v}$

For motion in one dimension: $P = Fv$. $\blacksquare$

Problem. A car of mass $1000\;\mathrm{kg}$ has an engine that produces a constant power of $40\;\mathrm{kW}$. The resistance to motion is $800\;\mathrm{N}$. Find the maximum speed of the car On a level road and the acceleration when the speed is $15\;\mathrm{m\,s^{-1}}$.

Maximum speed: At maximum speed, acceleration $= 0$So driving force $=$ resistance.

$P = Fv_{\max} = Rv_{\max}$

$40000 = 800v_{\max} \implies v_{\max} = 50\;\mathrm{m\,s^{-1}}$

At $v = 15\;\mathrm{m\,s^{-1}}$:

Driving force: $F = \dfrac{P}{v} = \dfrac{40000}{15} \approx 2666.7\;\mathrm{N}$.

$F - R = ma$

$2666.7 - 800 = 1000a \implies a = 1.867\;\mathrm{m\,s^{-2}}$

Problem. A car of mass $800\;\mathrm{kg}$ travels up a hill inclined at $\sin^{-1}(0.05)$ to The horizontal. The engine works at a constant $30\;\mathrm{kW}$ and the resistance is $300\;\mathrm{N}$. Find the maximum speed.

At maximum speed: driving force $= 300 + 800g\sin\alpha = 300 + 800(9.8)(0.05) = 300 + 392 = 692\;\mathrm{N}$.

$v_{\max} = \frac{P}{F} = \frac{30000}{692} \approx 43.4\;\mathrm{m\,s^{-1}}$

For an elastic spring (or string) that obeys Hooke’s Law, the tension (or thrust) is Proportional to the extension:

$T = kx$

Where $k$ is the stiffness (or spring constant) in $\mathrm{N\,m^{-1}}$ and $x$ is the extension In metres.

Alternatively, $T = \dfrac{\lambda x}{l}$ where $\lambda$ is the modulus of elasticity and $l$ Is the natural length.

The elastic potential energy (EPE) stored in a spring extended by $x$ from its natural length is:

$\boxed{\mathrm{EPE} = \frac{1}{2}kx^2 = \frac{\lambda x^2}{2l}}$

Derivation. The work done in extending the spring from $0$ to $x$:

$W = \int_0^x T\,dx = \int_0^x kx\,dx = \left[\frac{1}{2}kx^2\right]_0^x = \frac{1}{2}kx^2$

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

Problem. A light elastic string of natural length $1.2\;\mathrm{m}$ and modulus of elasticity $60\;\mathrm{N}$ has one end fixed and a particle of mass $2\;\mathrm{kg}$ attached to the other. The particle is released from rest at the point where the string is just taut. Find the maximum Extension and the maximum speed.

Let $x$ be the extension below the natural length position.

At maximum extension, speed $= 0$. By conservation of energy:

$\mathrm{GPE\ lost} = \mathrm{EPE\ gained}$

$mgx = \frac{\lambda x^2}{2l}$

$2gx = \frac{60x^2}{2(1.2)} = 25x^2$

$x(25x - 2g) = 0 \implies x = \frac{2g}{25} = \frac{19.6}{25} = 0.784\;\mathrm{m}$

Maximum speed occurs when acceleration $= 0$ (i.e., $mg = T$):

$2g = \frac{60x}{1.2} = 50x \implies x = \frac{2g}{50} = 0.392\;\mathrm{m}$

Energy conservation from start to this point:

$mgx = \frac{1}{2}mv^2 + \frac{\lambda x^2}{2l}$

$2(9.8)(0.392) = v^2 + \frac{60(0.392)^2}{2(1.2)}$

$7.683 = v^2 + \frac{60 \times 0.1537}{2.4} = v^2 + 3.842$

$v^2 = 3.841 \implies v \approx 1.96\;\mathrm{m\,s^{-1}}$

Problem. A particle of mass $3\;\mathrm{kg}$ is attached to two elastic strings. One string has Natural length $0.8\;\mathrm{m}$ and modulus $40\;\mathrm{N}$And is fixed at a point $A$. The other Has natural length $1.0\;\mathrm{m}$ and modulus $50\;\mathrm{N}$And is fixed at a point $B$. The Distance $AB$ is $3\;\mathrm{m}$. The particle hangs in equilibrium. Find the distance of the particle From $A$.

Let the particle be at distance $d$ from $A$ (and $3 - d$ from $B$).

Extension of string from $A$: $d - 0.8$ (if $d \gt 0.8$). Extension of string from $B$: $(3 - d) - 1.0 = 2 - d$ (if $d \lt 2$).

For equilibrium, both strings must be stretched, so $0.8 \lt d \lt 2$.

Resolving vertically (the particle hangs below the line $AB$):

$T_A + T_B = 3g$

$\frac{40(d - 0.8)}{0.8} + \frac{50(2 - d)}{1.0} = 29.4$

$50(d - 0.8) + 50(2 - d) = 29.4$

$50d - 40 + 100 - 50d = 29.4$

$60 = 29.4$

This is a contradiction, which means the particle does not hang directly below the line $AB$ in a Simple 1D configuration, or one of the strings is slack. If string $A$ is slack, then $d \leq 0.8$:

$T_B = 3g \implies \frac{50(2 - d)}{1.0} = 29.4 \implies 2 - d = 0.588 \implies d = 1.412$

But $d = 1.412 \gt 0.8$Contradicting $d \leq 0.8$. If string $B$ is slack ($d \geq 2$):

$T_A = 3g \implies 50(d - 0.8) = 29.4 \implies d = 1.388\;\mathrm{m}$

But $1.388 \lt 2$Contradiction. This problem needs a 2D treatment with the particle hanging below The line, with both strings at angles.

• Momentum: Energy and momentum are conserved
• Kinematics: Energy methods complement kinematics
• Forces: Work relates force to displacement