When a spring is compresssed by 3 cm the potential energy stored in it is U. When it is compressed further by 3cm, the increase in potential energy is

Options

A. 4U
B. U
C. 2U
D. 3U

Answer

To solve the problem of finding the increase in potential energy when a spring is compressed further, we can follow these steps:

### Step 1: Understand the Potential Energy Formula
The potential energy (U) stored in a spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the compression or extension of the spring from its equilibrium position.

### Step 2: Calculate the Initial Potential Energy (U1)
When the spring is compressed by 3 cm (which is 0.03 m), we can denote this potential energy as \( U_1 \):
\[ U_1 = \frac{1}{2} k (0.03)^2 = \frac{1}{2} k (9 \times 10^{-4}) = \frac{9}{2} k \times 10^{-4} \]
Given that \( U_1 = U \), we have:
\[ U = \frac{9}{2} k \times 10^{-4} \]

### Step 3: Calculate the Potential Energy After Further Compression (U2)
When the spring is compressed an additional 3 cm (making a total of 6 cm or 0.06 m), we denote this potential energy as \( U_2 \):
\[ U_2 = \frac{1}{2} k (0.06)^2 = \frac{1}{2} k (36 \times 10^{-4}) = 18 k \times 10^{-4} \]

### Step 4: Find the Increase in Potential Energy
The increase in potential energy when the spring is compressed from 3 cm to 6 cm is given by:
\[ \Delta U = U_2 - U_1 \]
Substituting the values we calculated:
\[ \Delta U = (18 k \times 10^{-4}) - \left(\frac{9}{2} k \times 10^{-4}\right) \]
To simplify this, we need a common denominator:
\[ \Delta U = \left(18 - \frac{9}{2}\right) k \times 10^{-4} = \left(\frac{36}{2} - \frac{9}{2}\right) k \times 10^{-4} = \frac{27}{2} k \times 10^{-4} \]

### Step 5: Relate the Increase in Potential Energy to U
From our earlier expression for \( U \):
\[ U = \frac{9}{2} k \times 10^{-4} \]
We can express \( k \times 10^{-4} \) in terms of \( U \):
\[ k \times 10^{-4} = \frac{2U}{9} \]
Substituting this back into our expression for \( \Delta U \):
\[ \Delta U = \frac{27}{2} \left(\frac{2U}{9}\right) = \frac{27U}{9} = 3U \]

### Final Answer
Thus, the increase in potential energy when the spring is compressed further by 3 cm is:
\[ \Delta U = 3U \]