answer this question---**Chart Description:**
* **Figure 1:** A diagram showing block Y positioned above block X. Block X is resting on a vertical spring, which is attached to a base (represented by a shaded bar).
* **Figure 2:** A diagram showing block Y stacked directly on top of block X. The combined blocks (Y above X) are resting on the same vertical spring, which is compressed compared to its likely equilibrium position without the blocks. The spring is attached to the same base.
**Question Stem:**
Block Y with mass $m_y$ falls onto and sticks to block X, which is attached to a vertical spring, as shown in Figure 1. A short time later, as shown in Figure 2, the blocks are momentarily at rest. At that moment, block Y exerts a force of magnitude $F_{\text{down}}$ on block X, and block X exerts a force of magnitude $F_{\text{up}}$ on block Y. Which of the following correctly relates $F_{\text{up}}$, $F_{\text{down}}$, and $m_y g$ at the instant shown in Figure 2?
**Options:**
(A) $(F_{\text{up}} = F_{\text{down}}) > m_y g$
(B) $(F_{\text{up}} = m_y g) > F_{\text{down}}$
(C) $m_y g > F_{\text{up}} > F_{\text{down}}$
(D) $F_{\text{up}} = F_{\text{down}} = m_y g$
视频信息
答案文本
视频字幕
In this physics problem, we have block Y with mass m_y falling onto block X, which is attached to a vertical spring. Block Y sticks to block X upon impact. We need to analyze the forces acting on the blocks when they are momentarily at rest after the collision.
Now let's analyze the forces acting on block Y when the blocks are momentarily at rest. Block Y experiences its weight m_y g acting downward, and a normal force F_up from block X acting upward. By Newton's third law, the force that block Y exerts on block X, F_down, is equal in magnitude to F_up.
Now we apply Newton's second law to block Y. The sum of forces equals mass times acceleration: F_up minus m_y g equals m_y times a_y. At the instant shown, the blocks are at the lowest point of their oscillation. While they are momentarily at rest, they have maximum upward acceleration toward the equilibrium position. Therefore, a_y is positive.
Now we can solve for the force relationship. From Newton's second law, F_up minus m_y g equals m_y times a_y. Since both m_y and a_y are positive, their product is positive. This means F_up minus m_y g is greater than zero, so F_up is greater than m_y g. Combined with Newton's third law that F_up equals F_down, we get that F_up equals F_down and both are greater than m_y g. This corresponds to answer choice A.
To summarize, this problem required three key physics concepts. First, Newton's second law relating force, mass, and acceleration. Second, Newton's third law stating that action and reaction forces are equal. Third, understanding simple harmonic motion, where at the lowest point the velocity is zero but acceleration is maximum upward. Combining these concepts leads us to the correct answer: F_up equals F_down and both are greater than m_y g, which is option A.