A block of mass 2 kg is dropped from a height of 40 cm on a spring whose force-constant `1960 N m^(-1)`. The maximum distance through which the spring is compressed by

Options

A. 10 cm
B. 1 cm
C. 20 cm
D. 5 cm

Explanation

To solve the problem of finding the maximum distance through which the spring is compressed when a block of mass 2 kg is dropped from a height of 40 cm onto a spring with a force constant of 1960 N/m, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
When the block is dropped, its gravitational potential energy will be converted into elastic potential energy of the spring at maximum compression.

### Step 2: Write the Expressions for Potential Energies
1. **Gravitational Potential Energy (GPE)** when the block is at height \( h \):
\[ \text{GPE} = mgh \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.4 \, \text{m} \) (height in meters)

2. **Elastic Potential Energy (EPE)** stored in the spring when compressed by distance \( x \):
\[ \text{EPE} = \frac{1}{2} k x^2 \]
where:
- \( k = 1960 \, \text{N/m} \) (spring constant)

### Step 3: Set Up the Energy Conservation Equation
At maximum compression, the gravitational potential energy lost will equal the elastic potential energy gained:
\[ mgh = \frac{1}{2} k x^2 \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ 2 \times 9.8 \times 0.4 = \frac{1}{2} \times 1960 \times x^2 \]
Calculating the left side:
\[ 2 \times 9.8 \times 0.4 = 7.84 \, \text{J} \]
So the equation becomes:
\[ 7.84 = \frac{1}{2} \times 1960 \times x^2 \]
This simplifies to:
\[ 7.84 = 980 x^2 \]

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

### Step 6: Take the Square Root to Find \( x \)
Taking the square root of both sides:
\[ x = \sqrt{0.008} = 0.0894 \, \text{m} = 8.94 \, \text{cm} \]

### Step 7: Round to the Nearest Option
The closest option to 8.94 cm is 10 cm. Therefore, the maximum distance the spring is compressed is approximately:
\[ \boxed{10 \, \text{cm}} \]