Cambridge Notes, Past Papers, Revision Questions

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{si}}^{\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

Weight acting on the body: \(\mathbf{W}=m\mathbf{g}\) (directed downwards).

Work done by the weight when the body moves a vertical distance \(h\) (upward positive) is
\[
W{\text{weight}} = \int{0}^{h} \mathbf{W}\!\cdot\!d\mathbf{s}
= \int_{0}^{h} (mg)(-\,\hat{y})\!\cdot\!\hat{y}\,dh' = -\,mg h .
\]

The external agent (e.g. a person lifting the object) does work
\[
W{\text{agent}} = -\,W{\text{weight}} = +\,mg h .
\]

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}= -\nabla\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)

GPE
A 2.0 kg crate is lifted 5.0 m vertically.
\[
U = mgh = (2.0)(9.81)(5.0)=98.1\ \text{J}.
\]

KE
A 0.50 kg ball is thrown upward with speed 10.0 m s\(^{-1}\).
\[
E_{K}= \tfrac12 m v^{2}= \tfrac12 (0.50)(10.0)^{2}=25.0\ \text{J}.
\]

Speed from GPE → KE
A 3.0 kg stone falls from rest from a height of 20.0 m (air resistance neglected).
\[
mg h = \tfrac12 m v^{2}\;\Longrightarrow\;
v = \sqrt{2gh}= \sqrt{2(9.81)(20.0)}=19.8\ \text{m s}^{-1}.
\]

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: \(Wf = -\muk m g d = -(0.20)(1.0)(9.81)(2.0)= -3.92\ \text{J}\).
3. Energy balance:
  \[
  Ui + Wf = E{K,f}\;\Longrightarrow\;E{K,f}=29.4-3.92=25.5\ \text{J}.
  \]
4. Final speed:
  \[
  v = \sqrt{2E_{K,f}/m}= \sqrt{2(25.5)/1.0}=7.14\ \text{m s}^{-1}.
  \]

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)

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 \(K_{\text{rot}}=\tfrac12 I\omega^{2}\).)

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.

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\(^{-2}\).

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.

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	E_K	\(\tfrac12 m v^{2}\)	J	E_K = 0.50 kg × (10.0 m s⁻¹)² = 25.0 J
Total Mechanical Energy	E_mech	U + E_K	J	E_mech = 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	W_nc	−μ_k m g d (kinetic friction)	J	W_f = −0.20 × 1.0 kg × 9.81 m s⁻² × 2.0 m = −3.92 J

recall and use EK = 21mv2