Forces and Newton’s Laws: Question 5

Worked solution

Part (a): Calculating the spring constant

The spring constant is defined by:

$k = \frac{F}{x}$

Using the results from $0$ to $8.0\text{ N}$, extension increases in equal steps of $1.0\text{ cm}$ for every $2.0\text{ N}$ added, so the data is directly proportional over this range. Taking, for example, $F = 8.0\text{ N}$ and $x = 4.0\text{ cm}$:

$k = \frac{8.0\text{ N}}{4.0\text{ cm}} = \boxed{2.0\text{ N/cm}}$

(Any pair of values from $0$ to $8.0\text{ N}$ gives the same result, e.g. $\dfrac{4.0}{2.0} = 2.0\text{ N/cm}$.)

Part (b): Identifying the limit of proportionality

In the proportional region, every extra $2.0\text{ N}$ of load produces exactly $1.0\text{ cm}$ more extension. If this pattern continued to $10.0\text{ N}$, the extension would be:

$4.0\text{ cm} + 1.0\text{ cm} = 5.0\text{ cm}$

However, the table shows the actual extension at $10.0\text{ N}$ is $6.0\text{ cm}$, which is larger than the $5.0\text{ cm}$ the proportional pattern predicts. This means the spring is no longer extending proportionally to the load at this point, so $\boxed{10.0\text{ N}}$ is beyond the limit of proportionality.

Part (c): Extension of the second spring

Rearrange $k = F/x$ to make extension the subject:

$k = \frac{F}{x} \quad\Rightarrow\quad x = \frac{F}{k}$

$x = \frac{9.0\text{ N}}{4.0\text{ N/cm}}$

$x = \boxed{2.25\text{ cm}}$

Part (d): Circular motion with a greater mass

The syllabus tells us, qualitatively, that an increased mass requires an increased force to keep the speed and radius of a circular path constant.

Here, the speed and the radius of the circle stay the same, but the mass of the ball increases. To keep a more massive ball moving at the same speed on the same circular path, the string must pull on it with a greater force directed towards the centre of the circle. (No formula is needed for this. It follows directly from the qualitative relationship between force, mass, speed and radius for circular motion.)

Final answers

• (a) Spring constant $=$ $2.0\text{ N/cm}$
• (b) $10.0\text{ N}$ is beyond the limit of proportionality, since its extension ($6.0\text{ cm}$) exceeds the $5.0\text{ cm}$ the proportional pattern predicts
• (c) Extension $=$ $2.25\text{ cm}$
• (d) The force provided by the string must increase