A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of te block. The block is givena sharp impulse from below so that it acquires an upward speed of 2.0 m/s. How high will it rise? Take `g=10 m/s^2`

Explanation

To solve the problem step by step, we will use the principles of energy conservation, specifically the relationship between kinetic energy and potential energy.

### Step 1: Identify the given data
- Mass of the block (m) = 5.0 kg
- Initial speed (v) = 2.0 m/s
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the initial kinetic energy (KE) of the block
The kinetic energy (KE) of the block can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 5.0 \, \text{kg} \times (2.0 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 5.0 \times 4 = 10 \, \text{J} \]

### Step 3: Set up the equation for potential energy (PE) at the maximum height
When the block rises to its maximum height, all the kinetic energy will be converted into potential energy (PE). The potential energy at height \( h \) is given by:
\[ PE = mgh \]
Setting the kinetic energy equal to the potential energy:
\[ KE = PE \]
\[ 10 \, \text{J} = 5.0 \, \text{kg} \times 10 \, \text{m/s}^2 \times h \]

### Step 4: Solve for height (h)
Rearranging the equation to solve for \( h \):
\[ h = \frac{10 \, \text{J}}{5.0 \, \text{kg} \times 10 \, \text{m/s}^2} \]
\[ h = \frac{10}{50} = 0.2 \, \text{m} \]

### Step 5: Convert height to centimeters
To express the height in centimeters:
\[ h = 0.2 \, \text{m} \times 100 \, \text{cm/m} = 20 \, \text{cm} \]

### Final Answer
The block will rise to a height of **20 cm**.

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