A spring with spring constaant k when compressed by 1 cm the PE stored is U. If it is further compressed by 3 cm, then change in its PE is

Options

A. 3 U
B. 9 U
C. 8 U
D. 15 U

Answer

To solve the problem step by step, we will use the formula for the potential energy (PE) stored in a spring, which is given by:

\[ PE = \frac{1}{2} k x^2 \]

where \( k \) is the spring constant and \( x \) is the compression or extension of the spring from its equilibrium position.

### Step-by-Step Solution:

1. **Initial Compression (1 cm)**:
- When the spring is compressed by 1 cm (which is 0.01 m), the potential energy \( U \) stored in the spring is:
\[ U = \frac{1}{2} k (0.01)^2 = \frac{1}{2} k \cdot 0.0001 = \frac{1}{20000} k \]

2. **Further Compression (Total 4 cm)**:
- If the spring is further compressed by 3 cm, the total compression becomes 4 cm (which is 0.04 m).
- The potential energy when the spring is compressed by 4 cm is:
\[ PE_{final} = \frac{1}{2} k (0.04)^2 = \frac{1}{2} k \cdot 0.0016 = \frac{1}{1250} k \]

3. **Relate PE to U**:
- We know from the first step that \( U = \frac{1}{20000} k \).
- To express \( PE_{final} \) in terms of \( U \):
\[ PE_{final} = \frac{1}{1250} k = \frac{1}{1250} k \cdot \frac{20000}{20000} = \frac{16}{20000} k = 16U \]

4. **Change in Potential Energy**:
- The change in potential energy \( \Delta PE \) when the spring is compressed from 1 cm to 4 cm is given by:
\[ \Delta PE = PE_{final} - U = 16U - U = 15U \]

### Final Answer:
The change in potential energy when the spring is further compressed by 3 cm is \( 15U \).

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