（1）奶奶所做的功：$W = Fs = 240\,\text{N} \times 6\,\text{m} = 1440\,\text{J}$
（2）爷爷的重力：$G_{\text{人}} = m_{\text{人}}g = 60\,\text{kg} \times 10\,\text{N/kg} = 600\,\text{N}$
由图可知，承担物重的绳子段数$n = 3，$不计绳重和摩擦，由$F = \frac{G_{\text{人}} + G_{\text{动}}}{n}$可得动滑轮（含座椅）的重力：$G_{\text{动}} = nF - G_{\text{人}} = 3 \times 240\,\text{N} - 600\,\text{N} = 120\,\text{N}$
动滑轮（含座椅）的质量：$m_{\text{动}} = \frac{G_{\text{动}}}{g} = \frac{120\,\text{N}}{10\,\text{N/kg}} = 12\,\text{kg}$
（3）爷爷上升的高度：$h = \frac{s}{n} = \frac{6\,\text{m}}{3} = 2\,\text{m}$
有用功：$W_{\text{有用}} = G_{\text{人}}h = 600\,\text{N} \times 2\,\text{m} = 1200\,\text{J}$
总功：$W_{\text{总}} = 1440\,\text{J}$
机械效率：$\eta = \frac{W_{\text{有用}}}{W_{\text{总}}} \times 100\% = \frac{1200\,\text{J}}{1440\,\text{J}} \times 100\% \approx 83.3\%$
