When a spring is stretched by 10 cm, the potential energy stored is E. When the spring is stretched by 10 cm more, the potential energy stored in the spring becomes A. 2E B. 4E C. 6E D. 10E

To unravel the relationship between the stretching of a spring and the potential energy it stores, we can utilize Hooke's Law, which states that the force needed to stretch or compress a spring is directly proportional to the distance it is stretched or compressed. This relationship also applies to the potential energy stored in the spring, which can be calculated using the formula:

Understanding the Energy Stored in a Spring

The formula for the potential energy (PE) stored in a spring is given by:

PE = (1/2) k x²

where k is the spring constant and x is the displacement from the spring's equilibrium position (the amount of stretch or compression).

Calculating the Potential Energy for Different Stretches

1. **When the spring is stretched by 10 cm**:

• Let’s denote this stretch as x₁ = 10 cm.
• The potential energy (E) stored in the spring can be expressed as:

E = (1/2) k (10 cm)²

2. **Now, when the spring is stretched by an additional 10 cm** (total stretch of 20 cm):

• Denote this total stretch as x₂ = 20 cm.
• The new potential energy (A) stored in the spring becomes:

A = (1/2) k (20 cm)²

• Expanding this, we get:

A = (1/2) k (400 cm²)

• So, we can simplify A:

A = 200 k

Relating the Two Energies

Now, we can compare E and A:

• From the first calculation, we have:

E = (1/2) k (100 cm²) = 50 k

• From the second calculation:

A = 200 k

To relate A to E, we can find the ratio:

A = 4E

Final Answer and Conclusion

Thus, when the spring is stretched by an additional 10 cm, the potential energy stored in the spring becomes 4E, which corresponds to option B.

This illustrates how potential energy increases with the square of the stretch, which is a crucial concept in understanding the behavior of springs and elastic materials in physics.