A spring stores 5J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its end is made to undergo small oscillations. If the block makes 5 oscillations each second what is the mass of the block?

Options

A. `0.16kg`
B. `1.6kg`
C. `16kg`
D. `0.016kg`

Explanation

To solve the problem step by step, we will follow the given information and apply relevant formulas.

### Step 1: Understand the Energy Stored in the Spring
The potential energy (PE) stored in a spring when it is stretched is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

### Step 2: Convert Units
We are given that the spring is stretched by 25 cm. We need to convert this to meters:
\[ x = 25 \text{ cm} = 0.25 \text{ m} \]

### Step 3: Set Up the Equation for Potential Energy
We know the potential energy stored in the spring is 5 J. Substituting the values into the potential energy formula:
\[ 5 = \frac{1}{2} k (0.25)^2 \]

### Step 4: Solve for the Spring Constant \( k \)
Now, we can solve for \( k \):
\[ 5 = \frac{1}{2} k (0.0625) \quad \text{(since } 0.25^2 = 0.0625\text{)} \]
\[ 5 = \frac{1}{2} k \cdot 0.0625 \]
\[ 10 = k \cdot 0.0625 \]
\[ k = \frac{10}{0.0625} = 160 \text{ N/m} \]

### Step 5: Determine the Frequency of Oscillation
The problem states that the block makes 5 oscillations each second. Thus, the frequency \( f \) is:
\[ f = 5 \text{ Hz} \]

### Step 6: Calculate the Time Period \( T \)
The time period \( T \) is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{5} = 0.2 \text{ seconds} \]

### Step 7: Use the Time Period Formula for a Spring-Mass System
The time period \( T \) for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the known values:
\[ 0.2 = 2\pi \sqrt{\frac{m}{160}} \]

### Step 8: Solve for Mass \( m \)
First, isolate the square root:
\[ \frac{0.2}{2\pi} = \sqrt{\frac{m}{160}} \]
Now square both sides:
\[ \left(\frac{0.2}{2\pi}\right)^2 = \frac{m}{160} \]
Calculating the left side:
\[ \frac{0.04}{4\pi^2} = \frac{m}{160} \]
Now multiply both sides by 160:
\[ m = 160 \cdot \frac{0.04}{4\pi^2} \]
\[ m = \frac{160 \cdot 0.04}{4\pi^2} = \frac{6.4}{\pi^2} \]
Using \( \pi^2 \approx 10 \):
\[ m \approx \frac{6.4}{10} = 0.64 \text{ kg} \]

### Final Answer
The mass of the block is approximately:
\[ m \approx 0.16 \text{ kg} \text{ or } 160 \text{ grams} \]