A small metallic sphere is suspended by a light spring of force constant k from the ceilling of a cage, which is accelerating uniformly by a force F in the upward direction. The ratio of mass of the cage to that of the sphere is 'n'. Find the potential energy stored in the spring.

Options

A. `(F^(2))/(2(n+1)^(2)k)`
B. `(F^(2))/((n+1)^(2)k)`
C. `(F^(2))/(2(n-1)^(2)k)`
D. `(F^(2))/((n-1)^(2)k)`

Explanation

To solve the problem of finding the potential energy stored in the spring when a small metallic sphere is suspended in a cage that is accelerating upward, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( m \) = mass of the sphere
- \( M \) = mass of the cage
- \( n \) = ratio of the mass of the cage to that of the sphere, i.e., \( n = \frac{M}{m} \)
- \( k \) = spring constant
- \( F \) = force causing the cage to accelerate
- \( a \) = acceleration of the cage
- \( g \) = acceleration due to gravity

### Step 2: Write the Equations of Motion
For the cage and the sphere, we can write the equations based on Newton's second law.

1. For the cage (upward direction):
\[ F - T' - Mg = Ma \tag{1} \]

2. For the sphere (downward direction):
\[ T' - mg = ma \tag{2} \]

Where \( T' \) is the tension in the spring.

### Step 3: Solve for Tension
From equation (1):
\[ T' = F - Ma - Mg \]

From equation (2):
\[ T' = mg + ma \]

Setting the two expressions for \( T' \) equal gives:
\[ F - Ma - Mg = mg + ma \]

### Step 4: Combine the Equations
Rearranging gives:
\[ F = (M + m)g + (M + m)a \]
This can be simplified to:
\[ F = (M + m)(g + a) \tag{3} \]

### Step 5: Express Tension in Terms of Spring Constant
Since \( T' = kx \) (where \( x \) is the extension of the spring), we can express \( x \) as:
\[ x = \frac{T'}{k} \]

Substituting \( T' \) from equation (2):
\[ x = \frac{mg + ma}{k} \]

### Step 6: Substitute Mass Ratio
Using the mass ratio \( n = \frac{M}{m} \), we can express \( M \) as \( M = nm \). Substituting this into our equations gives:
\[ F = (nm + m)(g + a) = m(n + 1)(g + a) \]

### Step 7: Calculate Potential Energy
The potential energy \( U \) stored in the spring is given by:
\[ U = \frac{1}{2} k x^2 \]

Substituting for \( x \):
\[ U = \frac{1}{2} k \left(\frac{mg + ma}{k}\right)^2 = \frac{1}{2} \frac{(mg + ma)^2}{k} \]

### Step 8: Final Expression
Substituting \( g + a \) in terms of \( F \):
\[ U = \frac{1}{2} \frac{(F/(n + 1))^2}{k} \]

Thus, the final expression for the potential energy stored in the spring is:
\[ U = \frac{F^2}{2k(n + 1)^2} \]

### Conclusion
The potential energy stored in the spring is:
\[ U = \frac{F^2}{2k(n + 1)^2} \]