Physics 9702/22 — May/June 2015

Background Concept

Work done $W$ by a force is the energy transferred when a force causes a displacement. For a constant force acting along the direction of motion,

$$W = Fs$$

where $F$ is force and $s$ is displacement. To find SI base units of any derived quantity, we rewrite it using definitions that involve SI base quantities (kg, m, s, A, K, mol, cd) and then simplify.

Understanding the Question

You are asked to start from the definition of work done and show that the SI base units of energy are $\text{kg m}^2\text{ s}^{-2}$. So you must:

• write the defining equation for work,
• express force in SI base units,
• multiply by displacement units, and simplify.

Approach

1. Use $W = Fs$.
2. Replace $F$ using Newton’s second law $F = ma$.
3. Use base units: $m$ (mass) in kg, $a$ (acceleration) in $\text{m s}^{-2}$, $s$ (displacement) in m.
4. Combine and simplify powers.

Step-by-Step Reasoning

Start with the definition:

$$W = Fs$$

So the units of work are units of force times units of displacement.

Force is defined by:

$$F = ma$$

Mass has units $\text{kg}$. Acceleration is change of velocity per time, so its units are $\text{m s}^{-2}$. Therefore:

$$[F] = \text{kg} \times \text{m s}^{-2} = \text{kg m s}^{-2}$$

Displacement $s$ has units m. Multiply:

$$[W] = (\text{kg m s}^{-2})(\text{m}) = \text{kg m}^2\text{ s}^{-2}$$

These are the SI base units of energy (the joule).

Key Takeaways

• Derived units come from definitions like $W = Fs$.
• Convert each quantity to base units, then simplify indices.
• Energy (joule) has base units $\text{kg m}^2\text{ s}^{-2}$.

Common Mistakes

• Using $W = \frac{1}{2}mv^2$ without showing it comes from definitions (the question specifically asks for the definition of work).
• Leaving force in newtons without converting to base units.
• Dropping a power of m, giving $\text{kg m s}^{-2}$ instead of $\text{kg m}^2\text{ s}^{-2}$.

Things to Be Careful About

• Distinguish mass symbol $m$ from metre (m); always rely on context.
• Write units using index notation (e.g. $\text{s}^{-2}$).
• Ensure you explicitly show the step $F = ma$ to reach base units.

Background Concept

Potential difference (p.d.) between two points in a circuit tells you how much energy is transferred when charge moves between those points. It is defined by the energy transferred per coulomb of charge.

Mathematically,

$$V = \frac{W}{Q}$$

where $V$ is potential difference, $W$ is work done / energy transferred, and $Q$ is charge.

Understanding the Question

The question asks you to define potential difference. This means you must give the standard physics meaning (not a description like “the voltage of a battery”). For 1 mark, the definition must include “work done/energy transferred per unit charge” (or equivalent wording).

Approach

State the definition in words and, if helpful, include the defining equation $V = W/Q$.

Step-by-Step Reasoning

• Consider moving a charge $Q$ between two points in a circuit.
• If energy $W$ is transferred (electrical energy to other forms, or supplied by a source), then the potential difference is defined as energy per charge.
• Therefore:

$$V = \frac{W}{Q}$$

Key Takeaways

• Potential difference measures energy transfer per unit charge.
• The key defining equation is $V = W/Q$.

Common Mistakes

• Defining it as “force per unit charge” (that is electric field strength, $E$).
• Saying “energy per unit current” or “current per unit voltage”.
• Forgetting “between two points” (p.d. is always between two locations).

Things to Be Careful About

• Use either “work done” or “energy transferred”; both are acceptable and equivalent here.
• Ensure “per unit charge” is explicit (e.g. “per coulomb”).

Background Concept

Resistance $R$ is defined by Ohm’s law (as a relationship between current and potential difference for a conductor at constant temperature):

$$R = \frac{V}{I}$$

To find SI base units of $R$, we need base units of $V$ and $I$. Current $I$ is already an SI base quantity with unit ampere (A). Potential difference is itself derived:

$$V = \frac{W}{Q}$$

where $W$ is energy transferred (joule, with base units $\text{kg m}^2\text{ s}^{-2}$) and $Q$ is charge. Charge is related to current by:

$$Q = It$$

so charge has units $\text{A s}$.

Understanding the Question

You must determine the SI base units of resistance and show working. That means the final unit must be in terms of base units only (kg, m, s, A). You are expected to start from $R = V/I$ and then break $V$ down into base units using definitions of p.d. and charge.

Approach

1. Start with $R = V/I$.
2. Replace $V$ with $W/Q$.
3. Replace $Q$ with $It$.
4. Substitute base units: $[W] = \text{kg m}^2\text{ s}^{-2}$, $[I] = \text{A}$, $[t] = \text{s}$.
5. Simplify powers of seconds and amperes carefully.

Step-by-Step Reasoning

Start from resistance:

$$R = \frac{V}{I}$$

So we need the units of $V$. From the definition of potential difference:

$$V = \frac{W}{Q}$$

Energy/work has base units (from part (a)):

$$[W] = \text{kg m}^2\text{ s}^{-2}$$

Now express charge $Q$ using current:

$$Q = It$$

So

$$[Q] = \text{A} \times \text{s} = \text{A s}$$

Therefore the units of potential difference are:

$$[V] = \frac{\text{kg m}^2\text{ s}^{-2}}{\text{A s}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}$$

Finally divide by current to get resistance:

$$[R] = \frac{[V]}{[I]} = \frac{\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}}{\text{A}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}$$

This is the SI base unit of resistance (the ohm, $\Omega$).

Key Takeaways

• Use $R = V/I$, then reduce $V$ using $V = W/Q$ and $Q = It$.
• Charge has units $\text{A s}$.
• Resistance in base units is $\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}$.

Common Mistakes

• Stopping at $\Omega$ instead of giving SI base units.
• Using $V = IR$ in a circular way without converting $V$ to base units.
• Forgetting that $Q = It$ so charge involves seconds; missing this leads to the wrong power of $s$.
• Writing $\text{kg m}^2\text{ s}^{-2}\text{ A}^{-2}$ (missing the extra $\text{s}^{-1}$ factor).

Things to Be Careful About

• Be consistent about symbols: $W$ here is work/energy, not watt.
• When dividing by $\text{A s}$, remember it subtracts indices: $\text{s}^{-2} / \text{s} = \text{s}^{-3}$.
• Final answer must be in base units only: kg, m, s, A (no coulomb, volt, newton, or ohm).