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 container of water is
d(cjx09nm r0 :g0p*: wvbutd tsitting
n28)tc p d:nqgepb2z2on a scale. Originally, the scale reads $M_{1}=45\,\mathrm{kg}$. A block of wood is suspended from a second scale; originally the scale read $M_{2}=12\,\mathrm{kg}$. The density of wood is $0.60\,\mathrm{g/cm^{3}}$; the density of the water is $1.00\,\mathrm{g/cm^{3}}$. The block of wood is lowered into the water until half of the block is beneath the surface.
What is the resulting reading on the scales?
A. $M_{1}=45\,\mathrm{kg}$ and $M_{2}=2\,\mathrm{kg}$.
B. $M_{1}=45\,\mathrm{kg}$ and $M_{2}=6\,\mathrm{kg}$.
C. $M_{1}=45\,\mathrm{kg}$ and $M_{2}=10\,\mathrm{kg}$.
D. $M_{1}=55\,\mathrm{kg}$ and $M_{2}=6\,\mathrm{kg}$.
E. $M_{1}=55\,\mathrm{kg}$ and $M_{2}=2\,\mathrm{kg}$.
参考答案: E
本题详细解析:
Use $g=10\,\mathrm{m/s^2}$. $\rho_{wood} = 600\,\mathrm{kg/m^3}$, $\rho_{water} = 1000\,\mathrm{kg/m^3}$.
1. **Find Buoyant Force ($F_B$)**:
Mass of wood $m_w = M_2 = 12\,\mathrm{kg}$.
Volume of wood $V_w = m_w / \rho_{wood} = 12\,\mathrm{kg} / 600\,\mathrm{kg/m^3} = 0.02\,\mathrm{m^3}$.
Volume submerged $V_{sub} = V_w / 2 = 0.01\,\mathrm{m^3}$.
Buoyant force $F_B = \rho_{water} \cdot g \cdot V_{sub} = (1000\,\mathrm{kg/m^3})(10\,\mathrm{m/s^2})(0.01\,\mathrm{m^3}) = 100\,\mathrm{N}$.
2. **Find new $M_2$**: Scale 2 reads the tension $T$ in the string. For the static block: $T + F_B = m_w g$.
$T = m_w g - F_B = (12\,\mathrm{kg})(10\,\mathrm{m/s^2}) - 100\,\mathrm{N} = 120\,\mathrm{N} - 100\,\mathrm{N} = 20\,\mathrm{N}$.
The scale reads mass: $M_2 = T/g = 20\,\mathrm{N} / 10\,\mathrm{m/s^2} = 2\,\mathrm{kg}$.
3. **Find new $M_1$**: By Newton's 3rd law, the buoyant force (up on the block) has a reaction force (down on the water). Scale 1 reads the original weight plus this reaction force.
$F_{scale1} = M_1 g + F_B = (45\,\mathrm{kg})(10\,\mathrm{m/s^2}) + 100\,\mathrm{N} = 450\,\mathrm{N} + 100\,\mathrm{N} = 550\,\mathrm{N}$.
The scale reads mass: $M_1 = F_{scale1}/g = 550\,\mathrm{N} / 10\,\mathrm{m/s^2} = 55\,\mathrm{kg}$.
