A block of mass `2kg` is propped from a heught of `40cm` on a spring where force constant is `1960Nm^(-1)` The maximum distance thought which the spring compressed by

Options

A. 5 cm
B. 15 cm
C. 20 cm
D. 10 cm

Explanation

To solve the problem step by step, we will use the principle of conservation of energy. The potential energy of the block at the height will be converted into the potential energy stored in the spring when it is compressed.

### Step 1: Identify the given values
- Mass of the block, \( m = 2 \, \text{kg} \)
- Height from which the block is dropped, \( h = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Spring constant, \( k = 1960 \, \text{N/m} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Write the expression for initial potential energy (PE_initial)
The initial potential energy of the block when it is at height \( h \) is given by:
\[ PE_{\text{initial}} = mgh \]
Substituting the values:
\[ PE_{\text{initial}} = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.4 \, \text{m} = 7.84 \, \text{J} \]

### Step 3: Write the expression for final potential energy (PE_final)
When the spring is compressed by a distance \( x_m \), the potential energy stored in the spring is given by:
\[ PE_{\text{final}} = \frac{1}{2} k x_m^2 \]

### Step 4: Apply the conservation of energy principle
According to the conservation of energy:
\[ PE_{\text{initial}} = PE_{\text{final}} \]
Thus,
\[ mgh = \frac{1}{2} k x_m^2 \]
Substituting the known values:
\[ 7.84 = \frac{1}{2} \times 1960 \times x_m^2 \]

### Step 5: Rearranging the equation to find \( x_m^2 \)
\[ 7.84 = 980 x_m^2 \]
\[ x_m^2 = \frac{7.84}{980} \]
\[ x_m^2 = 0.008 \]

### Step 6: Solve for \( x_m \)
Taking the square root of both sides:
\[ x_m = \sqrt{0.008} = 0.0894 \, \text{m} \]

### Step 7: Convert \( x_m \) to centimeters
To convert meters to centimeters, multiply by 100:
\[ x_m = 0.0894 \times 100 = 8.94 \, \text{cm} \]

### Step 8: Conclusion
The maximum distance through which the spring is compressed is approximately \( 8.94 \, \text{cm} \).