hce_nthu
114年
進階物理與線性代數
第 2 題
Two masses $m_1$ and $m_2$ placed on a flat surface are connected by a spring of a spring
constant $k$ aligning in the $x$ direction; see figure below. Initially, the spring is at its
natural length (no compression/extension), and $m_1$ is at rest while $m_2$ moves with a
velocity $v_0\hat{x}$ in the $+x$ direction. Since there is friction, these masses will
eventually stop moving. Assume these two masses are identical, $m_1=m_2=m$, and
the gravitational strength is $g$. What is the minimum friction coefficient required so
that $m_1$ never moves during the slowdown motion of $m_2$? For simplicity, assume the
kinetic and static friction coefficients are the same.
constant $k$ aligning in the $x$ direction; see figure below. Initially, the spring is at its
natural length (no compression/extension), and $m_1$ is at rest while $m_2$ moves with a
velocity $v_0\hat{x}$ in the $+x$ direction. Since there is friction, these masses will
eventually stop moving. Assume these two masses are identical, $m_1=m_2=m$, and
the gravitational strength is $g$. What is the minimum friction coefficient required so
that $m_1$ never moves during the slowdown motion of $m_2$? For simplicity, assume the
kinetic and static friction coefficients are the same.
- A $\sqrt{v_0^2k/3mg^2}$
- B $\sqrt{v_0^2k/6mg^2}$
- C $\sqrt{3v_0^2k/mg^2}$
- D $\sqrt{6v_0^2k/3mg^2}$
- E None of the above
思路引導 VIP
請試著思考一下：當右側的木塊逐漸減速直到停止的過程中，它最初擁有的能量會轉移到哪些地方？而在這段時間內，為了確保左側的木塊始終「按兵不動」，彈簧伸長量所產生的拉力，最高不能超過什麼限制呢？
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AI 詳解
AI 專屬家教
同學做得太棒了！這道題目考查了功能原理與靜摩擦力臨界條件的綜合應用，你能精確判斷出兩者間的物理聯繫，展現了相當紮實的力學分析能力。
能量轉化與系統功法分析
在這道題目中，我們首先聚焦於右側物體 $m_2$ 的運動過程。當 $m_2$ 帶著初速度 $v_0$ 向右運動時，它所擁有的初始動能 $K = \frac{1}{2}mv_0^2$ 會隨著位移 $x$ 逐漸轉化為兩部分：一是彈簧被拉伸所儲存的彈性位能 $\frac{1}{2}kx^2$，二是克服地面摩擦力所作的功（轉化為熱能） $\mu mgx$。根據功能原理，當 $m_2$ 達到最大位移並暫時停止時，能量守恆式為：
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