Cambridge Notes, Past Papers, Revision Questions

Work, Energy & Power – Kinetic and Gravitational Potential Energy

Learning Objectives

Recall the definition of work as a scalar (dot) product and interpret its sign.

State and apply the work‑energy theorem  ΔK = Σ W.

Derive the kinetic‑energy formula K = ½ mv² from the equations of motion.

Write the expression for gravitational potential energy U = m g h and explain the choice of reference level.

Use the principle of conservation of mechanical energy for systems with only conservative forces.

Define power (average and instantaneous) and calculate it in simple situations.

Explain efficiency and evaluate it when non‑conservative forces are present.

1. Work – scalar product

For a constant force \(\vec F\) acting through a straight‑line displacement \(\vec s\), the work done is

W = \vec F\!\cdot\!\vec s = F\,s\cos\theta

where \(\theta\) is the angle between the force and the displacement.

Scalar quantity: unit = joule (J) = N·m = kg·m²·s⁻².

Sign of work:
- Positive (\(\cos\theta>0\)) – force has a component in the direction of motion; energy is added to the system.
- Negative (\(\cos\theta<0\)) – force opposes the motion (e.g. friction, air resistance); energy is removed from the system.

2. Work–energy theorem

The total work done by all forces acting on a particle equals the change in its kinetic energy:

\Delta K = K{\text{final}}-K{\text{initial}} = \sum W

This is the work–energy theorem. It follows directly from Newton’s second law combined with the kinematic equations. The summation symbol emphasises that the work of every force (including gravity, normal reaction, friction, etc.) must be taken into account.

3. Derivation of the kinetic‑energy formula

Consider a particle of mass \(m\) moving in a straight line under a constant net force \(\vec F\). Choose the direction of motion as positive.

Newton’s second law: \(\displaystyle F = ma\).

For constant acceleration the kinematic relation is

v^{2}=u^{2}+2as,

where \(u\) and \(v\) are the initial and final speeds and \(s\) is the displacement.

Multiply by \(\dfrac{m}{2}\):

\frac12 m v^{2}= \frac12 m u^{2}+ m a s .

Replace \(ma\) by the net force \(F\):

\frac12 m v^{2}= \frac12 m u^{2}+ Fs .

The term \(Fs\) is the work done by the constant net force (see Section 1).

Re‑arrange:

\frac12 m v^{2}-\frac12 m u^{2}= Fs .

The left‑hand side is \(\Delta K\); the right‑hand side is the total work \(W\). Hence \(\Delta K = W\), confirming the work‑energy theorem.

If the particle starts from rest (\(u=0\)), the kinetic energy at speed \(v\) is

K = \frac12 m v^{2}.

This is the required kinetic‑energy expression.

4. Gravitational potential energy

In a uniform gravitational field the weight is \(\vec W = -mg\,\hat y\) (downward). For a vertical displacement \(h\) upward, the work done by gravity is

W_g = \vec W\!\cdot\!\vec s = (-mg)(-h)= -mg\,h .

Choosing a reference level where the potential energy is defined as zero (commonly the ground or the lowest point of the motion), the gravitational potential energy of the body is

U = m g h .

Key points:

The sign of \(h\) indicates the position relative to the chosen reference (above → \(h>0\), below → \(h<0\)).

Only the difference \(\Delta U = mg\Delta h\) is physically meaningful; the absolute zero can be placed anywhere for convenience.

5. Conservation of mechanical energy

If only conservative forces (e.g. gravity) do work, the total mechanical energy remains constant:

E_{\text{total}} = K + U = \text{constant}.

Example: a body released from rest at height \(h\) falls under gravity. The loss in potential energy equals the gain in kinetic energy:

mg h = \frac12 m v^{2}\quad\Longrightarrow\quad v = \sqrt{2gh}.

6. Power

Average power: \(\displaystyle P_{\text{avg}} = \frac{W}{t}\).

Instantaneous power (force parallel to velocity): \(\displaystyle P = Fv\).

Units: watt (W) = J s⁻¹ = N·m·s⁻¹ = kg·m²·s⁻³.

7. Efficiency

When non‑conservative forces (friction, air resistance, etc.) dissipate energy, the useful output is less than the input. Efficiency is defined as

\eta = \frac{\text{useful energy output}}{\text{energy input}}\times 100\%.

For a falling object, if the measured kinetic energy at the ground is \(K{\text{meas}}\) while the ideal value (no losses) would be \(K{\text{ideal}} = mgh\), then

\eta = \frac{K_{\text{meas}}}{mgh}\times 100\%.

8. Summary Table

Quantity	Symbol	Expression	Units
Work (constant force)	\(W\)	\(W = \vec F\!\cdot\!\vec s = Fs\cos\theta\)	J
Kinetic Energy	\(K\)	\(K = \dfrac12 mv^{2}\)	J
Gravitational Potential Energy	\(U\)	\(U = mgh\) (relative to chosen reference)	J
Mechanical Energy (conserved)	\(E_{\text{mech}}\)	\(E_{\text{mech}} = K + U\)	J
Average Power	\(P_{\text{avg}}\)	\(P_{\text{avg}} = \dfrac{W}{t}\)	W
Instantaneous Power	\(P\)	\(P = Fv\) (when \(\vec F\parallel\vec v\))	W
Efficiency	\(\eta\)	\(\displaystyle \eta = \frac{\text{useful }E}{\text{input }E}\times100\%\)	%

9. Worked Example – Falling ball

A 0.5 kg ball is released from rest at a height of 3.0 m above the ground. Find:

Its speed just before impact (neglect air resistance).

The kinetic energy at impact.

The efficiency if the measured kinetic energy is only 12 J.

Solution:

Speed from energy conservation: \(v = \sqrt{2gh} = \sqrt{2(9.8)(3.0)} = 7.67\ \text{m s}^{-1}\).

Kinetic energy: \(K = \frac12 m v^{2} = \frac12(0.5)(7.67)^{2}=14.7\ \text{J}\).

Ideal kinetic energy (no losses) = \(mgh = 0.5\times9.8\times3.0 = 14.7\ \text{J}\).
Efficiency: \(\displaystyle \eta = \frac{12}{14.7}\times100\% = 82\%.\)

10. Practice Questions

A 2.0 kg ball is dropped from rest from a height of 5.0 m. Calculate its kinetic energy just before it hits the ground. (Take \(g = 9.8\ \text{m s}^{-2}\).)

A 1500 kg car accelerates from 0 to 20 m s\(^{-1}\) on a level road. Determine the change in its kinetic energy and the average power if the acceleration takes 8 s.

Show that the work done by a constant horizontal force \(F\) over a distance \(d\) equals the increase in kinetic energy, using the steps of the derivation in Section 3.

A roller‑coaster car of mass 500 kg starts from rest at the top of a 25 m hill and reaches the bottom. Assuming no friction, find its speed at the bottom and its kinetic energy. Then, if friction dissipates 5 kJ during the descent, calculate the efficiency of the energy conversion.

Calculate the instantaneous power delivered by a 200 N force acting on a crate that moves with a speed of 3 m s\(^{-1}\) in the direction of the force.

11. Key Take‑aways

Work is the scalar product \(W=\vec F\!\cdot\!\vec s\); its sign tells whether energy is added (+) or removed (–) from the system.

The work–energy theorem links total work to the change in kinetic energy: \(\Delta K = \sum W\).

Kinetic energy of a body moving with speed \(v\) is \(K = \dfrac12 mv^{2}\).

Gravitational potential energy is \(U = mgh\); the zero level can be chosen arbitrarily.

When only conservative forces act, mechanical energy is conserved: \(K + U = \text{constant}\).

Power quantifies the rate of energy transfer: \(P_{\text{avg}} = W/t\) and \(P = Fv\) (instantaneous, when \(\vec F\parallel\vec v\)).

Efficiency expresses the proportion of input energy that appears as useful output in the presence of non‑conservative forces.

Suggested diagram: A mass \(m\) falling a vertical distance \(h\) under gravity. Show the weight \(mg\) acting downward, the upward‑positive \(y\)-axis, the initial speed \(u=0\), the final speed \(v\), and the displacement vector \(\vec s\).

derive, using the equations of motion, the formula for kinetic energy EK = 21mv2