以动滑轮和重物为研究对象，物体出水前后受力情况如图所示．

$3{{F}_{拉1}}+{{F}_{浮}}={{G}_{物}}+{{G}_{轮}}$    ①

$3{{F}_{拉2}}={{G}_{物}}+{{G}_{轮}}$             ②

${F}_{拉1}=\frac {{P}_{1}}{v} =\frac {480\text{W}}{0.3\text{m/s}} =1600\text{N}$

${F}_{拉2}=\frac {{P}_{2}}{v} =\frac {600\text{W}}{0.3\text{m/s}} =2000\text{N}$

由①、②、③、④式得

${{F}_{浮}}=3({{F}_{拉2}}-{{F}_{拉1}})=3\times (2000\text{N}-1600\text{N})=1200\text{N}$．

${V}_{排}={V}_{物}=\frac {{F}_{浮}}{{\rho  }_{水}g} =\frac {1200\text{N}}{1\times {10}^{3}\text{kg}/{\text{m}}^{3}\times 10\text{N/kg}} =0.12{\text{m}}^{3}$

重物在水中上升高度${h}_{1}=\frac {1}{3} {s}_{1}=\frac {1}{3} {vt}_{1}=\frac {1}{3} \times 0.3\text{m/s}\times 50\text{s}=5\text{m}$

重物的高度${h}_{2}=\frac {1}{3} {s}_{2}=\frac {1}{3} {vt}_{2}=\frac {1}{3} \times 0.3\text{m/s}\times 10\text{s}=1\text{m}$

重物的横截面积$S=\frac {{V}_{物}}{{h}_{2}} =\frac {0.12{\text{m}}^{3}}{1\text{m}} =0.12{\text{m}}^{2}$

打捞前重物上表面受水的压力${{F}_{压}}={{p}_{水}}S={{\rho}_{水}}g{{h}_{1}}S=1\times {{10}^{3}}\text{kg/}{{\text{m}}^{3}}\times 10\text{N/kg}\times 5\text{m}\times 0.12{{\text{m}}^{2}}=6\times {{10}^{3}}\text{N}$．

从②式得：${{G}_{物}}+{{G}_{轮}}=3{{F}_{拉2}}=3\times 2000\text{N}=6000\text{N}$

$\eta =\frac {{W}_{有用}}{{W}_{总}} =\frac {{G}_{物}h}{({G}_{物}+{G}_{动})h} $

$G_物=h (G_物+G_轮) =0.8\times6000 \text{N}=4800 \text{N}$

${\rho  }_{物}=\frac {{G}_{物}}{g{V}_{物}} =\frac {4800\text{N}}{10\text{N}/\text{kg}\times 0.12{\text{m}}^{3}} =4\times {10}^{3}\text{kg}/{\text{m}}^{3}$．