A 2kg block is dropped from a height of 0.4 m on a spring of force constant `1960Nm^(-1)`. The maximum compression of the spring is

Options

A. `10cm`
B. `20 cm`
C. `30cm `
D. `40cm`

Answer

To solve the problem of finding the maximum compression of a spring when a 2 kg block is dropped from a height of 0.4 m onto it, we can follow these steps:

### Step 1: Calculate the potential energy (PE) of the block at the height of 0.4 m.
The potential energy of the block when it is at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.4 \, \text{m} \) (height from which the block is dropped)

Calculating:
\[ PE = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.4 \, \text{m} = 7.848 \, \text{J} \]

### Step 2: Set the potential energy equal to the elastic potential energy (EPE) of the spring at maximum compression.
The elastic potential energy stored in the spring when compressed by a distance \( x \) is given by:
\[ EPE = \frac{1}{2} k x^2 \]
where:
- \( k = 1960 \, \text{N/m} \) (spring constant)
- \( x \) is the maximum compression of the spring.

Setting the potential energy equal to the elastic potential energy:
\[ mgh = \frac{1}{2} k x^2 \]
Substituting the values we have:
\[ 7.848 \, \text{J} = \frac{1}{2} \times 1960 \, \text{N/m} \times x^2 \]

### Step 3: Solve for \( x^2 \).
Rearranging the equation:
\[ x^2 = \frac{2 \times 7.848 \, \text{J}}{1960 \, \text{N/m}} \]
Calculating:
\[ x^2 = \frac{15.696}{1960} \approx 0.008 \, \text{m}^2 \]

### Step 4: Calculate \( x \).
Taking the square root:
\[ x = \sqrt{0.008} \approx 0.0894 \, \text{m} \]

### Step 5: Convert \( x \) to centimeters.
To convert meters to centimeters, multiply by 100:
\[ x \approx 0.0894 \, \text{m} \times 100 \approx 8.94 \, \text{cm} \]

### Final Answer:
The maximum compression of the spring is approximately **8.94 cm**.

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