A 2-kg block is dropped from a height of 0.4 m on a spring of force constant `k=1960 N`. Find the maximum distance the spring will be compressed. (Take `g = 9.8 ms^(-2)`)
[Hint : Consider conservation of energy]

Answer

To solve the problem of finding the maximum distance the spring will be compressed when a 2-kg block is dropped from a height of 0.4 m onto a spring with a force constant \( k = 1960 \, \text{N/m} \), we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the energies involved**:
- The block has gravitational potential energy when it is at height \( h = 0.4 \, \text{m} \).
- When the block compresses the spring, this potential energy is converted into the elastic potential energy of the spring.

2. **Calculate the gravitational potential energy (PE)**:
\[ PE = mgh \]
where:
- \( m = 2 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 0.4 \, \text{m} \)

Substituting the values:
\[ PE = 2 \times 9.8 \times 0.4 = 7.84 \, \text{J} \]

3. **Set up the equation for elastic potential energy (EPE)**:
The elastic potential energy stored in the spring when it is compressed by a distance \( x \) is given by:
\[ EPE = \frac{1}{2} k x^2 \]
where \( k = 1960 \, \text{N/m} \).

4. **Apply conservation of energy**:
At the maximum compression of the spring, all the gravitational potential energy will be converted into elastic potential energy:
\[ mgh = \frac{1}{2} k x^2 \]
Substituting the known values:
\[ 7.84 = \frac{1}{2} \times 1960 \times x^2 \]

5. **Solve for \( x^2 \)**:
Rearranging the equation gives:
\[ 7.84 = 980 x^2 \]
\[ x^2 = \frac{7.84}{980} \]
\[ x^2 = 0.008 \]

6. **Calculate \( x \)**:
Taking the square root of both sides:
\[ x = \sqrt{0.008} = 0.0894 \, \text{m} \]

7. **Final answer**:
The maximum distance the spring will be compressed is approximately \( 0.0894 \, \text{m} \) or \( 8.94 \, \text{cm} \).