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, we need to calculate the change in potential energy (PE) of a spring when it is compressed from an initial position to a final position. Here’s a step-by-step breakdown of the solution:

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

### Step 2: Calculate the initial potential energy when compressed by 1 cm
When the spring is compressed by 1 cm (which is 0.01 m), the potential energy stored is given as \( U \).
Using the formula:
\[ U = \frac{1}{2} k (0.01)^2 \]
This implies:
\[ U = \frac{1}{2} k (0.0001) = \frac{k}{20000} \]

### Step 3: Calculate the potential energy when compressed by 4 cm
If the spring is further compressed by an additional 3 cm, the total compression becomes 4 cm (1 cm + 3 cm = 4 cm). Now we need to calculate the potential energy at this new compression:
\[ U_{final} = \frac{1}{2} k (0.04)^2 \]
Calculating this gives:
\[ U_{final} = \frac{1}{2} k (0.0016) = \frac{k}{1250} \]

### Step 4: Calculate the change in potential energy
The change in potential energy (\( \Delta U \)) is the difference between the final potential energy and the initial potential energy:
\[ \Delta U = U_{final} - U_{initial} \]
Substituting the values we calculated:
\[ \Delta U = \frac{k}{1250} - \frac{k}{20000} \]

### Step 5: Find a common denominator and simplify
To subtract these fractions, we need a common denominator. The least common multiple of 1250 and 20000 is 20000:
\[ \Delta U = \frac{16k}{20000} - \frac{k}{20000} \]
\[ \Delta U = \frac{16k - k}{20000} = \frac{15k}{20000} \]

### Step 6: Relate \( k \) to \( U \)
From the initial condition, we have:
\[ U = \frac{1}{2} k (0.01)^2 \]
This means:
\[ k = \frac{20000U}{1} \]
Substituting this into the change in potential energy:
\[ \Delta U = \frac{15 \times 20000U}{20000} = 15U \]

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

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