Physics 9702/11 — May/June 2019

Background Concept

Many physical quantities are derived from the SI base units (kg, m, s, A, K, mol, cd). To identify a unit from its base-unit form, use the defining equations:

• Force:

$$F = ma$$

so units of force are

$$\text{N} = \text{kg}\,\text{m s}^{-2}$$

• Work / energy:

$$W = Fs$$

so units of work are

$$\text{J} = \text{N m}$$

• Power:

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

so units of power are

$$\text{W} = \text{J s}^{-1}$$

• Pressure:

$$p = \frac{F}{A}$$

so units of pressure are

$$\text{Pa} = \frac{\text{N}}{\text{m}^2}$$

Understanding the Question

You are given a base-unit expression:

$$\text{kg m}^2\ \text{s}^{-2}$$

and asked which named SI unit (joule, newton, pascal, watt) has exactly these base units.

Approach

Write each option in base units and compare with the target:

• Convert N using $F=ma$.
• Convert J using $W=Fs$ or $1\ \text{J}=1\ \text{N m}$.
• Convert Pa using $p=F/A$.
• Convert W using $P=W/t$.

Pick the one that matches $\text{kg m}^2\ \text{s}^{-2}$.

Step-by-Step Reasoning

Start from force:

$$1\ \text{N} = 1\ \text{kg m s}^{-2}$$

Energy (work) is force (\times) distance:

$$1\ \text{J} = 1\ \text{N m}$$

Substitute the base units of (\text{N}):

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

This exactly matches the given base units, so the correct option is joule.

(For checking distractors quickly)

• Newton: $\text{kg m s}^{-2}$ (missing one factor of m)
• Pascal: $\text{Pa} = \text{N m}^{-2} = \text{kg m}^{-1}\ \text{s}^{-2}$
• Watt: $\text{W} = \text{J s}^{-1} = \text{kg m}^2\ \text{s}^{-3}$

Key Takeaways

• Convert named SI units to base units using defining physics equations.
• $\text{J} = \text{N m} = \text{kg m}^2\ \text{s}^{-2}$.
• Power and pressure are common traps because they differ by factors of $\text{s}$ or $\text{m}$.

Common Mistakes

• Confusing newton with joule: forgetting the extra factor of distance (m) in energy.
• Writing watt as $\text{kg m}^2\ \text{s}^{-2}$ (it must include an extra $\text{s}^{-1}$).
• Getting pascal wrong by using $p = F \times A$ instead of $p = F/A$.

Things to Be Careful About

• Track powers carefully: pressure introduces $\text{m}^{-2}$, power introduces $\text{s}^{-1}$.
• Use negative indices consistently: $\text{s}^{-2}$ means divide by $\text{s}^2$.
• Always reduce systematically: start from base definitions ($F=ma$, $W=Fs$, $P=W/t$, $p=F/A$).