A block of mass 4 kg hangs from a spring constant `k=400(N)/(m)`. The block is pulled down through 15 cm below and released. What is its kinetic energy when the block is 10 above the equilibrium position.?

Options

A. `5J`
B. `2.5J`
C. `1J`
D. `1.9J`

Explanation

To solve the problem step by step, we will follow the principles of energy conservation in simple harmonic motion (SHM).

### Step 1: Identify the given values
- Mass of the block, \( m = 4 \, \text{kg} \)
- Spring constant, \( k = 400 \, \text{N/m} \)
- Initial displacement from the equilibrium position (amplitude), \( A = 15 \, \text{cm} = 0.15 \, \text{m} \)
- Displacement when calculating kinetic energy, \( x = -10 \, \text{cm} = -0.1 \, \text{m} \) (since it is above the equilibrium position)

### Step 2: Calculate the angular frequency (\( \omega \))
The angular frequency is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{400}{4}} = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 3: Calculate the total mechanical energy in the system
The total mechanical energy (\( E \)) in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
Substituting the values:
\[ E = \frac{1}{2} \times 400 \times (0.15)^2 = \frac{1}{2} \times 400 \times 0.0225 = 4.5 \, \text{J} \]

### Step 4: Calculate the potential energy (\( PE \)) when the block is at \( x = -0.1 \, \text{m} \)
The potential energy stored in the spring is given by:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \times 400 \times (-0.1)^2 = \frac{1}{2} \times 400 \times 0.01 = 2 \, \text{J} \]

### Step 5: Calculate the kinetic energy (\( KE \)) when the block is at \( x = -0.1 \, \text{m} \)
Using the conservation of energy, we know:
\[ E = KE + PE \]
Rearranging gives:
\[ KE = E - PE \]
Substituting the values:
\[ KE = 4.5 \, \text{J} - 2 \, \text{J} = 2.5 \, \text{J} \]

### Final Answer
The kinetic energy when the block is 10 cm above the equilibrium position is \( \boxed{2.5 \, \text{J}} \).

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