Force constants `K_(1) and K_(2)` of two springs are in the ratio `5:4`. They are stretched by same length. If potential energy stored in one spring is 25 J then potential energy stored in second spring is

Options

A. 25 J
B. 16 J
C. 100 J
D. 20 J

Answer

To solve the problem step by step, we will use the formula for potential energy stored in a spring and the given ratios of the spring constants.

### Step 1: Understand the relationship between the spring constants
We are given that the force constants (spring constants) of two springs, \( K_1 \) and \( K_2 \), are in the ratio \( 5:4 \). This can be expressed mathematically as:
\[ \frac{K_1}{K_2} = \frac{5}{4} \]
From this, we can express \( K_2 \) in terms of \( K_1 \):
\[ K_2 = \frac{4}{5} K_1 \]

### Step 2: Write the formula for potential energy
The potential energy \( PE \) stored in a spring when it is stretched by a length \( x \) is given by the formula:
\[ PE = \frac{1}{2} K x^2 \]
For the first spring, the potential energy is given as \( PE_1 = 25 \, J \):
\[ PE_1 = \frac{1}{2} K_1 x^2 = 25 \]

### Step 3: Solve for \( K_1 \)
From the equation for \( PE_1 \):
\[ \frac{1}{2} K_1 x^2 = 25 \]
Multiplying both sides by 2:
\[ K_1 x^2 = 50 \]
Thus, we can express \( K_1 \) as:
\[ K_1 = \frac{50}{x^2} \]

### Step 4: Find \( K_2 \) using the ratio
Now we can find \( K_2 \) using the relationship we established:
\[ K_2 = \frac{4}{5} K_1 = \frac{4}{5} \left(\frac{50}{x^2}\right) = \frac{200}{5x^2} = \frac{40}{x^2} \]

### Step 5: Calculate the potential energy of the second spring
Using the potential energy formula for the second spring:
\[ PE_2 = \frac{1}{2} K_2 x^2 \]
Substituting \( K_2 \):
\[ PE_2 = \frac{1}{2} \left(\frac{40}{x^2}\right) x^2 = \frac{1}{2} \cdot 40 = 20 \, J \]

### Conclusion
The potential energy stored in the second spring is:
\[ \boxed{20 \, J} \]