Physics 2007 JAMB Past Questions - Question 30: A string of length 4 m is extended by 0.02 m when a load off 0.4 kg is suspended at its end. What will be the length of the string when the applied force is 15 N? £ .

Correct Answer

D

Explanation

To solve this problem, you can use Hooke's Law, which relates the extension of a spring or string to the applied force. The formula for Hooke's Law is:

\[F = k \cdot \Delta L\]

Where:
- \(F\) is the force applied (in newtons, N).
- \(k\) is the spring constant (in N/m).
- \(\Delta L\) is the extension or change in length (in meters, m).

In this case, the string extends by 0.02 m (or 20 mm) when a 0.4 kg load is applied. You can calculate the spring constant (\(k\)) using this information.

First, calculate the weight of the load (force due to gravity):

\[F_{\text{load}} = m \cdot g\]

Where:
- \(m\) is the mass (0.4 kg).
- \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).

\[F_{\text{load}} = 0.4 \, \text{kg} \cdot 9.81 \, \text{m/s²} = 3.924 \, \text{N}\]

Now, you can use Hooke's Law to calculate the spring constant (\(k\)):

\[3.924 \, \text{N} = k \cdot 0.02 \, \text{m}\]

Solve for \(k\):

\[k = \frac{3.924 \, \text{N}}{0.02 \, \text{m}} = 196.2 \, \text{N/m}\]

Now that you have the spring constant, you can use it to calculate the extension (\(\Delta L\)) when a 15 N force is applied. Rearrange Hooke's Law to find \(\Delta L\):

\[\Delta L = \frac{F}{k}\]

Substitute the values:

\[\Delta L = \frac{15 \, \text{N}}{196.2 \, \text{N/m}} \approx 0.0764 \, \text{m}\]

So, when a 15 N force is applied, the string will extend by approximately 0.0764 meters. To find the final length of the string, add this extension to the original length of 4 meters:

\[L_{\text{final}} = 4 \, \text{m} + 0.0764 \, \text{m} \approx 4.0764 \, \text{m}\]

The length of the string when the applied force is 15 N is approximately 4.0764 meters.