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, we need to find the mass of the block attached to the spring, given the energy stored in the spring and the frequency of oscillation. Here’s a step-by-step solution:

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

### Step 2: Convert displacement to meters
The displacement given is 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 spring constant \( k \)
Rearranging the equation to solve for \( k \):
\[ 5 = \frac{1}{2} k (0.0625) \]
\[ 5 = 0.03125 k \]
\[ k = \frac{5}{0.03125} = 160 \, \text{N/m} \]

### Step 5: Use the formula for the period of oscillation
The period \( T \) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
We also know that frequency \( f \) is the reciprocal of the period:
\[ f = \frac{1}{T} \]
Given that the frequency is 5 oscillations per second:
\[ T = \frac{1}{5} \, \text{s} \]

### Step 6: Substitute \( T \) into the period formula
Substituting \( T \) into the period formula:
\[ \frac{1}{5} = 2\pi \sqrt{\frac{m}{160}} \]

### Step 7: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{5}\right)^2 = (2\pi)^2 \frac{m}{160} \]
\[ \frac{1}{25} = 4\pi^2 \frac{m}{160} \]

### Step 8: Solve for mass \( m \)
Rearranging the equation to solve for \( m \):
\[ m = \frac{160}{25 \cdot 4\pi^2} \]
Calculating \( 4\pi^2 \):
\[ 4\pi^2 \approx 39.478 \]
Thus:
\[ m = \frac{160}{25 \cdot 39.478} \]
\[ m \approx \frac{160}{986.95} \approx 0.162 \, \text{kg} \]

### Conclusion
The mass of the block is approximately:
\[ m \approx 0.16 \, \text{kg} \]

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