recall and use EK = 21mv2

Gravitational Potential Energy (GPE) and Kinetic Energy (KE)

1. Key Concepts (AO1)

• Energy is a scalar quantity – it can be stored, transferred or transformed.
• Mechanical energy = Gravitational PE + Kinetic energy.
• In the absence of non‑conservative forces (e.g. friction, air resistance) mechanical energy is conserved.
• A reference level must be chosen when assigning a numerical value to GPE; only the **difference** in GPE matters for the physics of a problem.

2. Work–Energy Theorem (AO2)

The work done by a force \(\mathbf{F}\) during a displacement \(\mathbf{s}\) is the dot‑product

\[ W = \int_{\mathbf{s_i}}^{\mathbf{s_f}} \mathbf{F}\!\cdot\!d\mathbf{s}\; . \]

The **work‑energy theorem** (as worded in the Cambridge syllabus) states:

“The net work done on a particle equals the change in its kinetic energy.”

\[ W_{\text{net}} = \Delta E_{\!K}=E_{K,f}-E_{K,i}. \]

3. Gravitational Potential Energy

3.1 Definition

For an object of mass \(m\) at a height \(h\) above a chosen reference level in a uniform gravitational field, the gravitational potential energy is

\[ U = m g h \]

• \(U\) – gravitational potential energy (J)
• \(m\) – mass (kg)
• \(g\) – magnitude of the gravitational field (≈ 9.81 m s\(^{-2}\) at sea level)
• \(h\) – vertical displacement from the reference level (m)

3.2 Derivation from the work‑energy theorem

1. Weight acting on the body: \(\mathbf{W}=m\mathbf{g}\) (directed downwards).
2. Work done **by the weight** when the body moves a vertical distance \(h\) (upward positive) is \[ W_{\text{weight}} = \int_{0}^{h} \mathbf{W}\!\cdot\!d\mathbf{s} = \int_{0}^{h} (mg)(-\,\hat{y})\!\cdot\!\hat{y}\,dh' = -\,mg h . \]
3. The external agent (e.g. a person lifting the object) does work \[ W_{\text{agent}} = -\,W_{\text{weight}} = +\,mg h . \]
4. From the work‑energy theorem the increase in kinetic energy of the object is zero (steady lift), so the work done by the agent appears as an increase in GPE: \[ \Delta U = W_{\text{agent}} = mg h . \]

3.3 Sign convention & reference level

• Choose any convenient level (ground, floor, centre of Earth, top of a hill, …).
• If the object is **above** the reference, \(h>0\) and \(U>0\).
• If the object is **below** the reference, \(h<0\) and \(U<0\). The magnitude still follows \(|U|=mg|h|\).
• Only \(\Delta U\) (difference between two positions) influences the motion.

3.4 A‑level extension – Gravitational potential \(\Phi\) and field \(\mathbf{g}\)

For a point mass or a spherical body the gravitational potential at a distance \(r\) from the centre is

\[ \Phi = -\frac{G M}{r}, \qquad \mathbf{g}= -abla\Phi = -\frac{G M}{r^{2}}\hat{r}, \]

so that the AS formula \(U=mgh\) is a special case of the more general expression \(U=m\Phi\) when \(r\) is large compared with the size of the Earth.

4. Kinetic Energy

4.1 Definition

\[ E_{K}= \frac12 m v^{2} \]

where \(v\) is the speed (m s\(^{-1}\)). KE is always non‑negative because it depends on the square of the speed.

4.2 Derivation from the work‑energy theorem (constant net force)

\[ \begin{aligned} W_{\text{net}} &= \int_{0}^{s}\mathbf{F}\!\cdot\!d\mathbf{s}=F s = m a s ,\\[4pt] \text{but } v^{2} &= v_{0}^{2}+2as \;\;(v_{0}=0) \;\Longrightarrow\; s=\frac{v^{2}}{2a}.\\[4pt] \Rightarrow\;W_{\text{net}} &= m a\left(\frac{v^{2}}{2a}\right)=\frac12 m v^{2}=E_{K}. \end{aligned} \]

5. Power (AO2)

• Instantaneous power is the rate at which work is done: \[ P = \frac{dW}{dt}= \mathbf{F}\!\cdot\!\mathbf{v}. \]
• Because work and kinetic energy are related, \[ P = \frac{dE}{dt}. \]
• Example: lifting a 10 kg load at a constant speed of 0.5 m s\(^{-1}\). \[ P = Fv = (mg)v = (10\times9.81)(0.5)=49.1\ \text{W}. \]

6. Conservation of Mechanical Energy

6.1 With only conservative forces

\[ E_{\text{mech,i}} = E_{\text{mech,f}},\qquad E_{\text{mech}} = U + E_{K}. \]

6.2 Including non‑conservative forces

\[ E_{\text{mech,i}} + W_{\text{nc}} = E_{\text{mech,f}}, \]

where \(W_{\text{nc}}\) is the total work done by non‑conservative forces (negative when they remove energy).

7. Worked Examples (AO2)

1. GPE
   A 2.0 kg crate is lifted 5.0 m vertically.
   \[ U = mgh = (2.0)(9.81)(5.0)=98.1\ \text{J}. \]
2. KE
   A 0.50 kg ball is thrown upward with speed 10.0 m s\(^{-1}\).
   \[ E_{K}= \tfrac12 m v^{2}= \tfrac12 (0.50)(10.0)^{2}=25.0\ \text{J}. \]
3. Speed from GPE → KE
   A 3.0 kg stone falls from rest from a height of 20.0 m (air resistance neglected).
   \[ mg h = \tfrac12 m v^{2}\;\Longrightarrow\; v = \sqrt{2gh}= \sqrt{2(9.81)(20.0)}=19.8\ \text{m s}^{-1}. \]
4. Non‑conservative work (friction)
   A 1.0 kg block slides down a frictionless 3.0 m‑high ramp, then across a horizontal surface with kinetic‑friction coefficient \(\mu_k=0.20\) over 2.0 m.
   
   1. Initial mechanical energy: \(U_i = mgh = (1.0)(9.81)(3.0)=29.4\ \text{J}\).
   2. Work of friction: \(W_f = -\mu_k m g d = -(0.20)(1.0)(9.81)(2.0)= -3.92\ \text{J}\).
   3. Energy balance: \[ U_i + W_f = E_{K,f}\;\Longrightarrow\;E_{K,f}=29.4-3.92=25.5\ \text{J}. \]
   4. Final speed: \[ v = \sqrt{2E_{K,f}/m}= \sqrt{2(25.5)/1.0}=7.14\ \text{m s}^{-1}. \]
5. Power while lifting
   A 12 kg box is lifted at a constant speed of 0.8 m s\(^{-1}\).
   \[ P = mgv = (12)(9.81)(0.8)=94.2\ \text{W}. \]

8. Practice Questions (AO2)

1. Derive the kinetic energy of a rotating solid cylinder of mass \(M\) and radius \(R\) using the work‑energy principle. (Hint: \(I=\tfrac12 MR^{2}\) and \(K_{\text{rot}}=\tfrac12 I\omega^{2}\).)
2. A roller‑coaster car of mass 500 kg starts from rest at the top of a 30 m hill, then travels through a loop where friction does 1.2 kJ of negative work. Calculate its speed at the top of the loop (height 15 m). State any assumptions.
3. Calculate the total mechanical energy of a 1.0 kg mass that is 10.0 m above the reference level and moving horizontally at 4.0 m s\(^{-1}\). Use \(g=9.81\) m s\(^{-2}\).
4. Explain how the sign convention for \(U=mgh\) changes if the reference level is chosen at the top of a hill rather than at ground level.
5. State the work‑energy theorem exactly as it appears in the Cambridge AS syllabus and give a short example illustrating its use.

9. Summary Table

Quantity	Symbol	Formula	Units	Typical Example
Gravitational Potential Energy	U	m g h	J	U = 2.0 kg × 9.81 m s⁻² × 5.0 m = 98.1 J
Kinetic Energy	EK	\(\tfrac12 m v^{2}\)	J	EK = 0.50 kg × (10.0 m s⁻¹)² = 25.0 J
Total Mechanical Energy	Emech	U + EK	J	Emech = 98.1 J + 25.0 J = 123.1 J
Power	P	F·v = dW/dt = dE/dt	W	P = (12 kg × 9.81 m s⁻²) × 0.8 m s⁻¹ = 94.2 W
Work of Non‑conservative Forces	Wnc	−μk m g d (kinetic friction)	J	Wf = −0.20 × 1.0 kg × 9.81 m s⁻² × 2.0 m = −3.92 J

10. Suggested Diagram