A body of mass 'm' is suspended to an ideal spring of force constant 'k'. The expected change in the position of the body due to an additional force 'F' acting vertically downwards is

Options

A. `(3F)/(2K)`
B. `(2F)/(K)`
C. `(5F)/(2K)`
D. `(4F)/(K)`

Explanation

To solve the problem, we need to determine the change in the position of a mass 'm' suspended from a spring with force constant 'k' when an additional downward force 'F' is applied. We will analyze the system using the principles of equilibrium and energy conservation.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- A mass 'm' is attached to a spring with spring constant 'k'.
- Initially, the mass causes the spring to stretch by a distance 'x' due to its weight (mg).

2. **Finding the Initial Displacement (x)**:
- At equilibrium, the force exerted by the spring (spring force) equals the weight of the mass.
- The spring force can be expressed as \( F_{\text{spring}} = kx \).
- Setting this equal to the weight of the mass, we have:
\[ kx = mg \]
- Solving for 'x', we get:
\[ x = \frac{mg}{k} \]

3. **Introducing the Additional Force (F)**:
- Now, when an additional force 'F' is applied downwards, the total downward force becomes \( mg + F \).
- The new displacement of the spring, denoted as \( x' \), can be found by setting the spring force equal to the new total weight:
\[ kx' = mg + F \]
- Solving for \( x' \), we have:
\[ x' = \frac{mg + F}{k} \]

4. **Calculating the Change in Position**:
- We need to find the change in position of the mass due to the additional force, which is given by \( x' - x \):
\[ x' - x = \left(\frac{mg + F}{k}\right) - \left(\frac{mg}{k}\right) \]
- Simplifying this expression:
\[ x' - x = \frac{mg + F - mg}{k} = \frac{F}{k} \]

5. **Final Result**:
- Therefore, the expected change in the position of the body due to the additional force 'F' acting vertically downwards is:
\[ \Delta x = x' - x = \frac{F}{k} \]

### Summary of the Solution:
The change in position of the mass due to the additional force is given by \( \Delta x = \frac{F}{k} \).

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