An object of mass 'm' is attached to a vertical fixed spring of spring constant 'k' and is allowed to fall under the gravity. What is the distance traversed by the object before being stopped ?

Options

A. mg/k
B. 2mg/k
C. mg/2k
D. none of these

Answer

To solve the problem of an object of mass 'm' attached to a vertical fixed spring of spring constant 'k' that is allowed to fall under gravity, we will apply the principle of conservation of energy. Here's the step-by-step solution:

### Step 1: Understand the Initial and Final Conditions
- Initially, the object is at rest at a height above the spring, and its initial potential energy is due to its height above the ground.
- When the object falls and comes to a stop, it has converted its gravitational potential energy into spring potential energy.

### Step 2: Write the Energy Conservation Equation
- The initial gravitational potential energy (PE_initial) when the object is at height 'x' is given by:
\[ PE_{\text{initial}} = mgh = mgx \]
where 'h' is the height fallen, which is equal to 'x' in this case.

- The potential energy stored in the spring (PE_spring) when it is stretched by 'x' is given by:
\[ PE_{\text{spring}} = \frac{1}{2} k x^2 \]

### Step 3: Set Up the Equation
- According to the conservation of energy, the total initial energy equals the total final energy:
\[ mgx = \frac{1}{2} k x^2 \]

### Step 4: Rearranging the Equation
- Rearranging the equation gives:
\[ \frac{1}{2} k x^2 - mgx = 0 \]
- Factoring out 'x' from the equation:
\[ x \left(\frac{1}{2} k x - mg\right) = 0 \]

### Step 5: Solve for 'x'
- This gives us two solutions:
1. \( x = 0 \) (the trivial solution where the object does not move)
2. \( \frac{1}{2} k x = mg \) which leads to:
\[ kx = 2mg \quad \Rightarrow \quad x = \frac{2mg}{k} \]

### Step 6: Conclusion
- The distance traversed by the object before being stopped by the spring is:
\[ x = \frac{2mg}{k} \]