Deformation of Solids: Question 1

Worked solution

Setting up the problem

Hooke’s law states that the force applied to a spring is directly proportional to its extension, within the limit of proportionality: $F = kx$ where $k$ is the spring constant. Rearranging for $k$: $k = \frac{F}{x}$

Before substituting, convert the extension to SI units (metres): $x = 4.0\text{ cm} = 4.0\times10^{-2}\text{ m} = 0.040\text{ m}$

Calculating the spring constant

$k = \frac{F}{x} = \frac{6.0}{0.040}$ $k = 150\text{ N m}^{-1}$

Why the other options are wrong

• A ($1.5\text{ N m}^{-1}$): this comes from failing to convert $4.0\text{ cm}$ to metres before dividing, i.e. computing $6.0 \div 4.0$.
• B ($24\text{ N m}^{-1}$): this comes from multiplying force by extension ($6.0 \times 0.040 \times 100$-type slip) instead of dividing, mixing up $k = Fx$ with $k = F/x$.
• D ($1500\text{ N m}^{-1}$): this comes from misplacing a power of ten in the conversion, treating $4.0\text{ cm}$ as $0.0040\text{ m}$ instead of $0.040\text{ m}$.

Final answer

• Spring constant $= \boxed{150}\text{ N m}^{-1}$, option C.