Question

A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground.

How much weight can be placed on the spring so that the end of the spring is 2 m above the ground?

96 N
288 N
384 N
480 N

Answer 1

To determine how much weight can be placed on the spring while ensuring that the end of the spring is 2 m above the ground, we can follow these steps:

1. **Calculate the distance the spring will stretch:**
The initial position of the end of the spring is 8 m above the ground. We want the end of the spring to be 2 m above the ground. Therefore, the spring must stretch by:
\[
\text{Stretch} = \text{Initial height} - \text{Final height} = 8 \, \text{m} - 2 \, \text{m} = 6 \, \text{m}
\]

2. **Use Hooke's Law to find the force exerted by the spring:**
According to Hooke's Law, the force exerted by a spring is given by:
\[
F = k \cdot x
\]
where:
- \( F \) is the force exerted by the spring,
- \( k \) is the spring constant (48 N/m),
- \( x \) is the stretch of the spring (6 m).
Plugging in the values:
\[
F = 48 \, \text{N/m} \cdot 6 \, \text{m} = 288 \, \text{N}
\]

3. **Interpret the result:**
The force exerted by the spring when it is stretched by 6 m is 288 N. This means that a weight of 288 N can be placed on the spring to achieve the desired position of the end of the spring 2 m above the ground.

**Final answer: 288 N**