Physics 9702/21 — October/November 2013

Background Concept

Energy is measured in joules ($\text{J}$). A joule can be defined from mechanical work:

$$W = Fs$$

where $F$ is force and $s$ is distance moved in the direction of the force.

The unit of force is the newton ($\text{N}$), and from Newton’s second law:

$$F = ma$$

so

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

Understanding the Question

You are asked to show the SI base units of the elastic potential energy $E$. Even though the context is an elastic wire, the unit of energy is the same for all forms of energy.

Approach

Convert energy to base units by:

1. using $E = Fs$ so $[E] = [F][s]$
2. writing $\text{N}$ in base units using $F = ma$
3. multiplying by $\text{m}$.

Step-by-Step Reasoning

1. Start from work/energy:

$$E = Fs \quad \Rightarrow \quad [E] = [F][s]$$

1. Write force in base units:

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

1. Multiply by distance $s$ (unit $\text{m}$):

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

This matches the expected base units of the joule.

Key Takeaways

• Energy (joule) in base units is $\text{kg m}^2\,\text{s}^{-2}$.
• A reliable route is $E = Fs$ with $F = ma$.

Common Mistakes

• Stopping at $\text{N m}$ without converting $\text{N}$ to base units.
• Writing $\text{kg m s}^{-1}$ (missing a power of $s$).
• Confusing mass ($\text{kg}$) with weight ($\text{N}$).

Things to Be Careful About

• Keep track of indices carefully when multiplying units: $\text{m}\times\text{m} = \text{m}^2$.
• Use negative indices for seconds: $\text{s}^{-2}$, not “per second squared”.

Background Concept

Dimensional (unit) analysis uses the fact that an equation must be homogeneous: both sides must have the same dimensions/units.

If

$$E = C\rho^2 g^2 A l_0^3$$

then the units of $C$ must “fix” the units of the product $\rho^2 g^2 A l_0^3$ so that the final units match those of energy $E$.

Key base-unit expressions:

• density:

$$\rho = \frac{m}{V} \Rightarrow [\rho]=\frac{\text{kg}}{\text{m}^3}=\text{kg m}^{-3}$$

• acceleration:

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

• area:

$$[A]=\text{m}^2$$

• length:

$$[l_0]=\text{m}$$

• energy:

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

Understanding the Question

You are given a formula for elastic potential energy $E$ in terms of density $\rho$, gravitational field strength $g$, cross-sectional area $A$, original length $l_0$, and a constant $C$.

The task is not to calculate a number but to determine what SI base units $C$ must have for the equation to be dimensionally consistent.

Approach

1. Write each quantity ($E,\rho,g,A,l_0$) in SI base units.
2. Find the combined units of $\rho^2 g^2 A l_0^3$ by applying powers and multiplying.
3. Rearrange unit-wise:

$$[C]=\frac{[E]}{[\rho^2 g^2 A l_0^3]}$$

Step-by-Step Reasoning

1. List base units:

$$[E]=\text{kg m}^2\,\text{s}^{-2}$$ $$[\rho]=\text{kg m}^{-3} \Rightarrow [\rho^2]=\text{kg}^2\text{m}^{-6}$$ $$[g]=\text{m s}^{-2} \Rightarrow [g^2]=\text{m}^2\text{s}^{-4}$$ $$[A]=\text{m}^2$$ $$[l_0]=\text{m} \Rightarrow [l_0^3]=\text{m}^3$$

1. Multiply the units on the right-hand side excluding $C$:

• kilograms: only from $\rho^2$, so $\text{kg}^2$
• metres: add powers: $-6 + 2 + 2 + 3 = +1$, so $\text{m}^1$
• seconds: only from $g^2$, so $\text{s}^{-4}$

Therefore:

$$[\rho^2 g^2 A l_0^3]=\text{kg}^2\text{m}\,\text{s}^{-4}$$

1. Solve for $[C]$:

$$[C]=\frac{\text{kg m}^2\,\text{s}^{-2}}{\text{kg}^2\text{m}\,\text{s}^{-4}}$$

Divide by subtracting indices:

• $\text{kg}^{1-2} = \text{kg}^{-1}$
• $\text{m}^{2-1} = \text{m}^{1}$
• $\text{s}^{-2-(-4)} = \text{s}^{2}$

So:

$$[C]=\text{kg}^{-1}\text{m}\,\text{s}^2$$

Key Takeaways

• Dimensional homogeneity lets you find the units of an unknown constant.
• When multiplying quantities, add indices; when dividing, subtract indices.
• Writing everything in base units avoids mistakes.

Common Mistakes

• Using $[\rho]=\text{kg m}^{-2}$ (forgetting density is per volume, $\text{m}^3$).
• Forgetting to square $\rho$ and $g$.
• Adding indices incorrectly for the metre powers (this is the most common arithmetic slip).
• Giving the unit as $\text{kg m s}^2$ (missing the negative power on $\text{kg}$).

Things to Be Careful About

• Check each quantity’s physical meaning before writing units: $A$ is area ($\text{m}^2$), not $\text{m}$.
• Keep the base-unit form consistent (use $\text{kg}$, $\text{m}$, $\text{s}$ only).
• A quick sanity check: since $\rho^2 g^2 A l_0^3$ has $\text{kg}^2$ but $E$ only has $\text{kg}$, $C$ must include $\text{kg}^{-1}$, which matches the result.