An object of mass 1 kg falls on a vertical spring from a height of 2 m, and spring gets compressed. If spring constant $K=200~N/m$, then the compression in the spring is (take $g=10~m/s^{2}$)

Question

Accepted Answer

To determine the compression in the spring when an object of mass 1 kg falls on it from a height of 2 m, we need to apply the principle of conservation of energy. The potential energy of the object due to its height will be converted into the elastic potential energy of the spring when it is compressed.

1. **Initial Potential Energy (PE) of the object:**
   $$
   PE = mgh = 1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 2 \, \text{m} = 20 \, \text{J}
   $$

2. **Elastic Potential Energy (EPE) of the spring when compressed:**
   $$
   EPE = \frac{1}{2} K x^2
   $$
   where $ K = 200 \, \text{N/m} $ is the spring constant and $ x $ is the compression in the spring.

3. **Applying conservation of energy:**
   $$
   mgh = \frac{1}{2} K x^2
   $$
   $$
   20 = \frac{1}{2} \times 200 \times x^2
   $$
   $$
   20 = 100x^2
   $$
   $$
   x^2 = \frac{20}{100} = 0.2
   $$
   $$
   x = \sqrt{0.2} = \frac{1}{\sqrt{5}} \, \text{m}
   $$

Thus, the compression in the spring is $\frac{1}{\sqrt{5}} \, \text{m}$, which corresponds to Option 3.

**Explanation for why other options are incorrect:**

- **Option 1 (0.5 m):** If the compression were 0.5 m, the elastic potential energy would be:
  $$
  EPE = \frac{1}{2} \times 200 \times (0.5)^2 = 25 \, \text{J}
  $$
  This exceeds the initial potential energy of 20 J, which is not possible, thus making this option incorrect.

- **Option 2 (0.4 m):** If the compression were 0.4 m, the elastic potential energy would be:
  $$
  EPE = \frac{1}{2} \times 200 \times (0.4)^2 = 16 \, \text{J}
  $$
  This is less than the initial potential energy of 20 J, indicating that not all the energy is accounted for, thus making this option incorrect.

- **Option 4 ($\frac{1}{5} \, \text{m}$):** If the compression were $\frac{1}{5} \, \text{m}$, the elastic potential energy would be:
  $$
  EPE = \frac{1}{2} \times 200 \times \left(\frac{1}{5}\right)^2 = 4 \, \text{J}
  $$
  This is significantly less than the initial potential energy of 20 J, indicating that a large portion of the energy is unaccounted for, thus making this option incorrect.

Therefore, the correct answer is indeed Option 3, $\frac{1}{\sqrt{5}} \, \text{m}$.