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A block is attached to a compressed, horizontal spring. A person is initially holding the block in place at position \( x=-\mathrm{A} \) as shown in the figure. The person then slowly moves the block at a constant speed through position \( x=0 \), at which point the spring is at its relaxed length, and then to position \( x=+\mathrm{A} \) Which of the following correctly describes how the potential energy in the spring-block system changes as the block is moved from \( x=-\mathrm{A} \) to \( x=+\mathrm{A} \) ? (A) The potential energy decreases throughout the motion (B) The potential energy increases throughout the motion (C) The potential energy decreases and then increases (D) The potential energy increases and then decreases (D)

Ask by Park West. in the United States
Feb 26,2025

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The potential energy decreases as the block moves from \( x=-\mathrm{A} \) to \( x=0 \) and then increases as it moves from \( x=0 \) to \( x=+\mathrm{A} \).

Solución

We measure the spring’s potential energy with U = (½)k(x)², where x is the displacement from the equilibrium (relaxed) position. In this setup, x = 0 corresponds to the spring’s relaxed, unstretched (or uncompressed) length. Initially, at x = –A, the spring is compressed. Its potential energy is U = (½)kA². As the person moves the block slowly to x = 0, the compression is undone and the potential energy decreases until it reaches U = 0 at x = 0. Continuing to move the block to x = +A stretches the spring; therefore, the potential energy increases from 0 at x = 0 back up to U = (½)kA² at x = +A. Thus, the potential energy decreases from x = –A to x = 0 and then increases from x = 0 to x = +A. The correct answer is (C) The potential energy decreases and then increases.

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As the block is moved from \( x=-\mathrm{A} \) (compressed spring) to \( x=+\mathrm{A} \) (stretched spring), the potential energy of the spring-block system reflects the energy stored due to the displacement of the spring from its equilibrium position. Initially, at \( x=-\mathrm{A} \), the potential energy is at its maximum (compression) and then gradually decreases to zero as the block passes through \( x=0 \) where the spring is relaxed. However, as it continues to move towards \( x=+\mathrm{A} \), potential energy starts to increase again, reaching its maximum when the spring is fully stretched. Thus, the correct answer would be (C) – the potential energy decreases and then increases. This elegant behavior illustrates the beauty of energy conservation in spring systems, where energy continually transfers between potential and kinetic forms as the block interacts with the spring. Remember that the spring's potential energy is given by \( U = \frac{1}{2} k x^2 \), which helps visualize how energy varies with displacement!

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