When a body of mass `1.0 kg` is suspended from a certain light spring hanging vertically, its length increases by `5cm`. By suspending `2.0kg` block to the spring and if the block is pulled through `10cm` and released, the maximum velocity of it in `m//s is (g = 10 m//s^(2))`

Options

A. `0.5`
B. `1`
C. `2`
D. `4`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the spring constant (K)

When a mass of 1.0 kg is suspended from the spring, it stretches the spring by 5 cm. We can use Hooke's Law, which states that the force exerted by the spring is proportional to the displacement (x) from its equilibrium position:

\[ F = Kx \]

At equilibrium, the weight of the mass (mg) is equal to the restoring force of the spring (Kx):

\[ mg = Kx_1 \]

Where:
- \( m = 1.0 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( x_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)

Substituting the values:

\[ 1.0 \times 10 = K \times 0.05 \]

\[ 10 = K \times 0.05 \]

Now, solve for K:

\[ K = \frac{10}{0.05} = 200 \, \text{N/m} \]

### Step 2: Calculate the maximum velocity (Vmax) when a 2.0 kg block is suspended

When a 2.0 kg block is suspended from the spring and pulled down by 10 cm (0.1 m), we need to find the maximum velocity. The formula for maximum velocity in a spring-mass system is given by:

\[ V_{max} = A \omega \]

Where:
- \( A = 0.1 \, \text{m} \) (the amplitude)
- \( \omega = \sqrt{\frac{K}{m}} \)

Now, we need to calculate \( \omega \):

Substituting the values of K and m:

\[ \omega = \sqrt{\frac{200}{2}} = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 3: Substitute values to find Vmax

Now substitute \( A \) and \( \omega \) into the maximum velocity formula:

\[ V_{max} = 0.1 \times 10 = 1 \, \text{m/s} \]

### Final Answer

The maximum velocity of the block when it is pulled through 10 cm and released is:

\[ V_{max} = 1 \, \text{m/s} \]

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