本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A spring system is set up as follows: a u6e 3 sm0/ 7du+ui gkewde/(baplatform with a weight o 3lwbj 9;n4i1/ oqzzysf $10 N$ is on top of two springs, each with spring constant $75 N/m$. On top of the platform is a third spring with spring constant $75 N/m$. If a ball with a weight of $5.0 N$ is then fastened to the top of the third spring and then slowly lowered, by how much does the height of the spring system change? 
A. $0.033 m$
B. $0.067 m$
C. $0.100 m$
D. $0.133 m$
E. $0.600 m$
参考答案: C
本题详细解析:
The two bottom springs are in parallel, so their equivalent spring coi vh b3) nj(.i d*o8-,klv6fydnz(r0t b fvqq8nstant is $k_{bottom} = k_1 + k_2 = 75 N/m + 75 N/m = 150 N/m$. The top spring is in series with this combination. The added weight of the ball $W_{ball} = 5.0 N$ compresses both the top spring and the bottom combination.
1. Compression of the top spring (due to $W_{ball}$): $x_{top} = W_{ball} / k_{top} = 5.0 N / 75 N/m \approx 0.0667 m$.
2. Compression of the bottom springs (due to $W_{ball}$): $x_{bottom} = W_{ball} / k_{bottom} = 5.0 N / 150 N/m \approx 0.0333 m$.
The total change in height is the sum of these compressions: $\Delta x = x_{top} + x_{bottom} = 0.0667 m + 0.0333 m = 0.100 m$.
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