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The elastic potential energy of a spring...

The elastic potential energy of a spring

A

Increases only when it is stretched

B

Decreases only when it is stretched

C

Decreases only when it is compressed

D

Increases whether stretched or compressed

Text Solution

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The correct Answer is:
To find the elastic potential energy of a spring, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Spring Force**: The force exerted by a spring is given by Hooke's Law, which states that the force \( F \) exerted by a spring is proportional to the displacement \( x \) from its equilibrium position. This can be expressed as: \[ F = -kx \] where \( k \) is the spring constant. **Hint**: Remember that the negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement. 2. **Work Done on the Spring**: When we stretch or compress the spring, we do work against the spring force. The work done \( W \) in stretching or compressing the spring from an initial position \( x = 0 \) to a position \( x \) is given by: \[ W = \int_0^x F \, dx = \int_0^x kx \, dx \] **Hint**: The work done is the integral of the force over the displacement. 3. **Calculating the Work Done**: Evaluating the integral: \[ W = \int_0^x kx \, dx = \left[ \frac{1}{2} kx^2 \right]_0^x = \frac{1}{2} kx^2 \] This work done is stored as elastic potential energy \( U \) in the spring. **Hint**: The limits of integration help you find the total work done from the initial to the final position. 4. **Elastic Potential Energy**: The elastic potential energy \( U \) stored in the spring when it is either stretched or compressed by a distance \( x \) is given by: \[ U = \frac{1}{2} kx^2 \] This formula shows that the potential energy increases as the displacement \( x \) increases, whether the spring is stretched or compressed. **Hint**: The formula indicates that potential energy is always positive when \( k \) and \( x \) are positive. ### Conclusion: The elastic potential energy of a spring is given by the formula: \[ U = \frac{1}{2} kx^2 \] This energy increases when the spring is either stretched or compressed.
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Knowledge Check

  • A 242 × 10^(4)kg freight car moving along a horizontal rail road spur track at 7.2 km/hour strikes a bumper whose coil springs experiences a maximum compression of 30 cm in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15 cm is

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