A string of length 5cm is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [g = 10ms$^{-2}$]

Analysis

To solve this problem, we use Hooke’s Law, which states that the extension of an elastic material is directly proportional to the force applied, provided the elastic limit is not exceeded. The formula is:

Where:

• $F$ = Force (in Newtons)
• $k$ = Force constant (stiffness of the string)
• $e$ = Extension (in meters)

Step-by-Step Solution

Step 1: Calculate the force applied in the first scenario. The load is given as $0.8\text{ kg}$. To find the force ($F_1$), we use $F = mg$:

Step 2: Find the force constant ($k$) of the string. Using the first scenario where $F_1 = 8\text{ N}$ and extension $e_1 = 0.04\text{ m}$:

Step 3: Calculate the extension for the second force. Now, we apply a force $F_2 = 16\text{ N}$ to the same string (the constant $k$ remains the same):

Conclusion

The string will extend by $0.08\text{ m}$ when a force of $16\text{ N}$ is applied.

Exam Tip: In questions involving Hooke's Law, you can also use the ratio method to save time:

Since $F_2$ ($16\text{ N}$) is exactly double $F_1$ ($8\text{ N}$), the extension $e_2$ must also be double the original extension $e_1$ ($0.04\text{ m} \times 2 = 0.08\text{ m}$). This shortcut is very useful for JAMB where time management is key!