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

Answer

To solve the problem, we need to calculate the change in potential energy (PE) of a spring when it is compressed from an initial position to a further compressed position. Let's break this down step by step.

### Step 1: Understand the potential energy formula for a spring
The potential energy (U) stored in a spring when compressed or stretched is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where:
- \( U \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the compression or extension from the natural length.

### Step 2: Calculate the potential energy when compressed by 1 cm
When the spring is compressed by 1 cm (which is 0.01 m), the potential energy is given as:
\[ U = \frac{1}{2} k (0.01)^2 \]
This is the initial potential energy, which we denote as \( U \).

### Step 3: Calculate the potential energy when further compressed by 3 cm
If the spring is further compressed by an additional 3 cm, the total compression becomes:
\[ x = 1 \text{ cm} + 3 \text{ cm} = 4 \text{ cm} = 0.04 \text{ m} \]
Now, we calculate the potential energy at this new compression:
\[ U_b = \frac{1}{2} k (0.04)^2 \]

### Step 4: Express \( U_b \) in terms of \( U \)
Now, substituting \( U_b \):
\[ U_b = \frac{1}{2} k (0.04)^2 = \frac{1}{2} k (0.04 \times 0.04) = \frac{1}{2} k (0.0016) \]
We can relate this to our initial potential energy \( U \):
\[ U = \frac{1}{2} k (0.01)^2 = \frac{1}{2} k (0.0001) \]
Now, we can express \( U_b \) in terms of \( U \):
\[ U_b = \frac{1.6}{0.1} U = 16U \]

### Step 5: Calculate the change in potential energy
The change in potential energy (\( \Delta U \)) when the spring is compressed from 1 cm to 4 cm is:
\[ \Delta U = U_b - U \]
Substituting the values we found:
\[ \Delta U = 16U - U = 15U \]

### Final Answer
The change in potential energy when the spring is compressed from 1 cm to 4 cm is:
\[ \Delta U = 15U \]