To solve the problem, we need to determine the amplitude of oscillation of a block attached to a spring when the elevator's upward acceleration ceases. Here’s a step-by-step solution:
### Step 1: Identify the Forces Acting on the Block
When the elevator is accelerating upward, two forces act on the block:
1. The gravitational force (weight) acting downward: \( F_g = mg \)
2. The pseudo force due to the elevator's upward acceleration: \( F_p = ma \)
Where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( a = 5 \, \text{m/s}^2 \) (upward acceleration of the elevator)
### Step 2: Calculate the Total Downward Force
The total downward force when the elevator is accelerating is:
\[
F_{\text{total}} = mg + ma = m(g + a)
\]
Substituting the values:
\[
F_{\text{total}} = 2 \, \text{kg} \times (10 \, \text{m/s}^2 + 5 \, \text{m/s}^2) = 2 \, \text{kg} \times 15 \, \text{m/s}^2 = 30 \, \text{N}
\]
### Step 3: Relate the Forces to the Spring Constant
The spring force when the elevator is accelerating is equal to the total downward force:
\[
F_{\text{spring}} = kx_1
\]
Where:
- \( k = 400 \, \text{N/m} \) (spring constant)
- \( x_1 \) is the extension of the spring when the elevator is accelerating.
Setting the forces equal gives:
\[
kx_1 = 30 \, \text{N}
\]
Solving for \( x_1 \):
\[
x_1 = \frac{30 \, \text{N}}{400 \, \text{N/m}} = 0.075 \, \text{m}
\]
### Step 4: Analyze the Situation When the Elevator Stops Accelerating
At \( t = 0 \), the elevator stops accelerating, and the only force acting on the block is its weight:
\[
F_{\text{gravity}} = mg
\]
The spring force now balances only the weight of the block:
\[
kx_2 = mg
\]
Where \( x_2 \) is the new extension of the spring after the acceleration stops. Substituting the values:
\[
kx_2 = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N}
\]
Solving for \( x_2 \):
\[
x_2 = \frac{20 \, \text{N}}{400 \, \text{N/m}} = 0.05 \, \text{m}
\]
### Step 5: Calculate the Amplitude of Oscillation
The amplitude of oscillation is the difference between the two extensions:
\[
A = x_1 - x_2
\]
Substituting the values:
\[
A = 0.075 \, \text{m} - 0.05 \, \text{m} = 0.025 \, \text{m}
\]
Converting to centimeters:
\[
A = 0.025 \, \text{m} = 2.5 \, \text{cm}
\]
### Final Answer
The amplitude of the oscillation is **2.5 cm**.
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