Physics 9702/22 — May/June 2014

Background Concept

SI base units are the fundamental units (e.g. $\text{kg}$, $\text{m}$, $\text{s}$) used to build up derived units. For mechanics:

• Force is defined by Newton’s second law:

$$F = ma$$

so its base units are $\text{kg m s}^{-2}$.

• Work done (energy transferred) by a constant force along the direction of motion is:

$$W = Fs$$

• Power is the rate of energy transfer:

$$P = \frac{W}{t}$$

The key skill is to substitute each quantity into base units and simplify.

Understanding the Question

You are asked to show (i.e. derive, not just state) the SI base units of power. The final answer must be expressed only using base units, so it should end in a combination of $\text{kg}$, $\text{m}$ and $\text{s}$ with powers.

Approach

1. Start with a defining equation for power: $P = W/t$.
2. Replace $W$ using $W = Fs$.
3. Replace $F$ using $F = ma$ and write base units for $m$ and $a$.
4. Simplify indices of $\text{m}$ and $\text{s}$.

Step-by-Step Reasoning

From the definition of power:

$$P = \frac{W}{t}$$

Work done is force times distance:

$$W = Fs$$

Substitute into the power equation:

$$P = \frac{Fs}{t}$$

Now convert force into base units using $F = ma$.

• Mass $m$ has unit $\text{kg}$.
• Acceleration $a$ has unit $\text{m s}^{-2}$.

So

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

Substitute units into $P = Fs/t$:

$$[P] = \frac{(\text{kg m s}^{-2})(\text{m})}{\text{s}} = \text{kg m}^2\,\text{s}^{-3}$$

This is the required SI base-unit form.

Key Takeaways

• Use definitions ($P = W/t$, $W = Fs$, $F = ma$) to reduce derived quantities to base units.
• Combine units by treating them like algebraic symbols with indices.
• Power in base units is $\text{kg m}^2\,\text{s}^{-3}$.

Common Mistakes

• Using $P = IV$ (still valid, but then you must correctly reduce $\text{A}$ and $\text{V}$ to base units; many students get that wrong).
• Stopping at $\text{J s}^{-1}$ without converting joules to base units.
• Arithmetic error with indices (e.g. ending with $\text{s}^{-2}$ instead of $\text{s}^{-3}$).

Things to Be Careful About

• Ensure the final expression contains only base units, not derived units like N or J.
• Keep track of the division by time: dividing by $\text{s}$ reduces the power of $\text{s}$ by 1.

Background Concept

Dimensional analysis uses the fact that any valid physical equation must be homogeneous: both sides must have the same dimensions (and therefore the same units).

Here $Q$ is thermal energy, measured in joules (J), and $Q/t$ is the rate of energy flow, i.e. power (W). In SI base units:

• $1\ \text{W} = 1\ \text{J s}^{-1} = \text{kg m}^2\,\text{s}^{-3}$
• Area $A$ has units $\text{m}^2$
• Thickness $x$ has units $\text{m}$
• Temperature difference $T$ has units kelvin (K)

A constant like $C$ often “soaks up” whatever units are needed to make the equation consistent.

Understanding the Question

You are given a formula for thermal power flow:

$$\frac{Q}{t} = \frac{CAT}{x}$$

You must find the SI base units of $C$. That means rearrange to make $C$ the subject and then substitute the base units of all the other quantities.

Approach

1. Rearrange the equation to isolate $C$.
2. Replace each symbol with its SI base units:
   • $Q/t$ as watts in base units,
   • $A$, $T$, $x$ in their SI units.
3. Simplify, cancelling powers of metres and collecting indices.

Step-by-Step Reasoning

Start with

$$\frac{Q}{t} = \frac{CAT}{x}$$

Make $C$ the subject by multiplying both sides by $x$ and dividing by $AT$:

$$C = \frac{Q}{t}\,\frac{x}{AT}$$

Now substitute units.

• $Q/t$ is power, so $[Q/t] = \text{W} = \text{kg m}^2\,\text{s}^{-3}$.
• $[x] = \text{m}$.
• $[A] = \text{m}^2$.
• $[T] = \text{K}$.

So

$$[C] = (\text{kg m}^2\,\text{s}^{-3})\times \frac{\text{m}}{\text{m}^2\,\text{K}}$$

Combine the metre powers: $\text{m}^2 \times \text{m} = \text{m}^3$, then divide by $\text{m}^2$ leaves $\text{m}^1$.

Thus

$$[C] = \text{kg m s}^{-3}\,\text{K}^{-1}$$

These are SI base units (kg, m, s, K).

Key Takeaways

• Rearranging first makes it clear which units belong to the constant.
• $Q/t$ is power, so it carries the base units $\text{kg m}^2\,\text{s}^{-3}$.
• Carefully cancel $\text{m}$ powers and keep temperature in kelvin.

Common Mistakes

• Treating $T$ as degrees Celsius: temperature differences can be in $^\circ\text{C}$ numerically, but the SI unit is K, and the unit symbol should be K.
• Using $Q$ (joules) instead of $Q/t$ (watts) when substituting units.
• Algebra slip in rearrangement (e.g. putting $x$ in the denominator instead of numerator).

Things to Be Careful About

• The question uses $T$ for temperature difference, but time is also sometimes written as $t$; keep them distinct.
• Ensure the final answer is in base units only; do not leave it as $\text{W m}^{-1}\,\text{K}^{-1}$ unless explicitly asked (here it asks for base units).
• Check homogeneity: once you have $[C]$, substituting back should give $\text{kg m}^2\,\text{s}^{-3}$ for $Q/t$.