A 2kg block hangs without vibrating at the bottom end of a spring with a force constant of 400N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of 5m/s2 when the acceleration suddenly ceases at time t=0 and the car moves upward with constant speed (g=10m/s2 The amplitude of the oscillation is

Answer

To solve the problem step by step, we need to analyze the forces acting on the block when the elevator is accelerating and then when it moves with constant speed.

### Step 1: Understand the forces acting on the block
When the elevator is accelerating upwards, the effective force acting on the block is the sum of the gravitational force and the force due to the elevator's acceleration. The gravitational force acting on the block is given by:
\[ F_g = m \cdot g \]
where \( m = 2 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the gravitational force
\[ F_g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Calculate the effective force when the elevator is accelerating
The effective force acting on the block when the elevator is accelerating upwards with an acceleration \( a = 5 \, \text{m/s}^2 \) is:
\[ F_{\text{effective}} = F_g + F_a = m \cdot g + m \cdot a \]
\[ F_{\text{effective}} = 20 \, \text{N} + 2 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 20 \, \text{N} + 10 \, \text{N} = 30 \, \text{N} \]

### Step 4: Relate the effective force to the spring constant
According to Hooke's law, the force exerted by the spring is:
\[ F_s = k \cdot x_0 \]
where \( k = 400 \, \text{N/m} \) and \( x_0 \) is the extension of the spring when the block is at rest.

Setting the effective force equal to the spring force:
\[ k \cdot x_0 = F_{\text{effective}} \]
\[ 400 \, \text{N/m} \cdot x_0 = 30 \, \text{N} \]

### Step 5: Solve for \( x_0 \)
\[ x_0 = \frac{30 \, \text{N}}{400 \, \text{N/m}} = 0.075 \, \text{m} \]

### Step 6: Determine the amplitude of oscillation
The amplitude of oscillation is equal to the extension of the spring when the elevator was accelerating upwards. Thus, the amplitude \( A \) is:
\[ A = x_0 = 0.075 \, \text{m} = 7.5 \, \text{cm} \]

### Final Answer
The amplitude of the oscillation is \( 7.5 \, \text{cm} \).

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