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1. (2023·兰州)一辆燃油汽车在平直公路上匀速直线行驶了108km,所用时间为1h,消耗汽油10kg,汽车在行驶过程中所受阻力恒为1380N.(汽油的热值q汽油$= 4.6×10^7J/kg)$
(1)求汽车行驶的速度.
(2)求牵引力做功的功率.
(3)求汽车发动机的效率.
(1)求汽车行驶的速度.
(2)求牵引力做功的功率.
(3)求汽车发动机的效率.
答案
(1) $ v = \frac{s}{t} = \frac{108 \, \text{km}}{1 \, \text{h}} = 108 \, \text{km/h} $
(2) $ F = f = 1380 \, \text{N} $，$ W = Fs = 1380 \, \text{N} \times 108 \times 10^3 \, \text{m} = 1.4904 \times 10^8 \, \text{J} $，$ P = \frac{W}{t} = \frac{1.4904 \times 10^8 \, \text{J}}{1 \times 3600 \, \text{s}} = 4.14 \times 10^4 \, \text{W} $
(3) $ Q_{\text{放}} = m q_{\text{汽油}} = 10 \, \text{kg} \times 4.6 \times 10^7 \, \text{J/kg} = 4.6 \times 10^8 \, \text{J} $，$ \eta = \frac{W}{Q_{\text{放}}} \times 100\% = \frac{1.4904 \times 10^8 \, \text{J}}{4.6 \times 10^8 \, \text{J}} \times 100\% = 32.4\% $
解析：(1) 汽车行驶的速度 $ v = \frac{s}{t} = \frac{108 \, \text{km}}{1 \, \text{h}} = 108 \, \text{km/h} $。(2) 汽车匀速直线行驶，受力平衡，根据二力平衡条件可知，牵引力 $ F = f = 1380 \, \text{N} $，牵引力做的功 $ W = Fs = 1380 \, \text{N} \times 108 \times 10^3 \, \text{m} = 1.4904 \times 10^8 \, \text{J} $，牵引力的功率 $ P = \frac{W}{t} = \frac{1.4904 \times 10^8 \, \text{J}}{1 \times 3600 \, \text{s}} = 4.14 \times 10^4 \, \text{W} $。(3) 消耗的汽油完全燃烧放出的热量 $ Q_{\text{放}} = m q_{\text{汽油}} = 10 \, \text{kg} \times 4.6 \times 10^7 \, \text{J/kg} = 4.6 \times 10^8 \, \text{J} $，汽车发动机的效率 $ \eta = \frac{W}{Q_{\text{放}}} \times 100\% = \frac{1.4904 \times 10^8 \, \text{J}}{4.6 \times 10^8 \, \text{J}} \times 100\% = 32.4\% $。
(2) $ F = f = 1380 \, \text{N} $，$ W = Fs = 1380 \, \text{N} \times 108 \times 10^3 \, \text{m} = 1.4904 \times 10^8 \, \text{J} $，$ P = \frac{W}{t} = \frac{1.4904 \times 10^8 \, \text{J}}{1 \times 3600 \, \text{s}} = 4.14 \times 10^4 \, \text{W} $
(3) $ Q_{\text{放}} = m q_{\text{汽油}} = 10 \, \text{kg} \times 4.6 \times 10^7 \, \text{J/kg} = 4.6 \times 10^8 \, \text{J} $，$ \eta = \frac{W}{Q_{\text{放}}} \times 100\% = \frac{1.4904 \times 10^8 \, \text{J}}{4.6 \times 10^8 \, \text{J}} \times 100\% = 32.4\% $
解析：(1) 汽车行驶的速度 $ v = \frac{s}{t} = \frac{108 \, \text{km}}{1 \, \text{h}} = 108 \, \text{km/h} $。(2) 汽车匀速直线行驶，受力平衡，根据二力平衡条件可知，牵引力 $ F = f = 1380 \, \text{N} $，牵引力做的功 $ W = Fs = 1380 \, \text{N} \times 108 \times 10^3 \, \text{m} = 1.4904 \times 10^8 \, \text{J} $，牵引力的功率 $ P = \frac{W}{t} = \frac{1.4904 \times 10^8 \, \text{J}}{1 \times 3600 \, \text{s}} = 4.14 \times 10^4 \, \text{W} $。(3) 消耗的汽油完全燃烧放出的热量 $ Q_{\text{放}} = m q_{\text{汽油}} = 10 \, \text{kg} \times 4.6 \times 10^7 \, \text{J/kg} = 4.6 \times 10^8 \, \text{J} $，汽车发动机的效率 $ \eta = \frac{W}{Q_{\text{放}}} \times 100\% = \frac{1.4904 \times 10^8 \, \text{J}}{4.6 \times 10^8 \, \text{J}} \times 100\% = 32.4\% $。
2. (2024·陕西)氢能因其低碳、高热值、来源广泛等特点在新能源领域备受青睐.某款氢内燃机汽车的总质量为1.8t,车轮与地面的总接触面积为$0.1m^2.$某次测试中,该车匀速直线行驶6km用时5min,牵引力做功的功率恒为36kW,发动机的效率为60%.(g取10N/kg,氢燃料的热值为$1.4×10^8J/kg,$假设氢燃料完全燃烧)
(1)该车静止在水平地面上时对地面的压强是多少?
(2)本次测试中,该车行驶的速度是多少?受到的牵引力是多少?
(3)本次测试中,消耗氢燃料的质量是多少?(保留2位小数)
(1)该车静止在水平地面上时对地面的压强是多少?
(2)本次测试中,该车行驶的速度是多少?受到的牵引力是多少?
(3)本次测试中,消耗氢燃料的质量是多少?(保留2位小数)
答案
(1) $ p = \frac{F}{S} = \frac{G}{S} = \frac{mg}{S} = \frac{1.8 \times 10^3 \, \text{kg} \times 10 \, \text{N/kg}}{0.1 \, \text{m}^2} = 1.8 \times 10^5 \, \text{Pa} $
(2) $ v = \frac{s}{t} = \frac{6 \times 10^3 \, \text{m}}{5 \times 60 \, \text{s}} = 20 \, \text{m/s} $，$ F = \frac{P}{v} = \frac{36 \times 10^3 \, \text{W}}{20 \, \text{m/s}} = 1800 \, \text{N} $
(3) $ W = Pt = 36 \times 10^3 \, \text{W} \times 5 \times 60 \, \text{s} = 1.08 \times 10^7 \, \text{J} $，$ Q_{\text{放}} = \frac{W}{\eta} = \frac{1.08 \times 10^7 \, \text{J}}{60\%} = 1.8 \times 10^7 \, \text{J} $，$ m_{\text{氢}} = \frac{Q_{\text{放}}}{q_{\text{氢}}} = \frac{1.8 \times 10^7 \, \text{J}}{1.4 \times 10^8 \, \text{J/kg}} \approx 0.13 \, \text{kg} $
解析：(1) 该车静止在水平地面上时对地面的压强 $ p = \frac{F}{S} = \frac{G}{S} = \frac{mg}{S} = \frac{1.8 \times 10^3 \, \text{kg} \times 10 \, \text{N/kg}}{0.1 \, \text{m}^2} = 1.8 \times 10^5 \, \text{Pa} $。(2) 本次测试中，该车匀速直线行驶 $ 6 \, \text{km} $ 用时 $ 5 \, \text{min} $，则该车行驶的速度 $ v = \frac{s}{t} = \frac{6 \times 10^3 \, \text{m}}{5 \times 60 \, \text{s}} = 20 \, \text{m/s} $；牵引力做功的功率恒为 $ 36 \, \text{kW} $，则该车受到的牵引力 $ F = \frac{P}{v} = \frac{36 \times 10^3 \, \text{W}}{20 \, \text{m/s}} = 1800 \, \text{N} $。(3) 本次测试中，牵引力做的功 $ W = Pt = 36 \times 10^3 \, \text{W} \times 5 \times 60 \, \text{s} = 1.08 \times 10^7 \, \text{J} $，氢燃料完全燃烧放出的热量 $ Q_{\text{放}} = \frac{W}{\eta} = \frac{1.08 \times 10^7 \, \text{J}}{60\%} = 1.8 \times 10^7 \, \text{J} $，消耗氢燃料的质量 $ m_{\text{氢}} = \frac{Q_{\text{放}}}{q_{\text{氢}}} = \frac{1.8 \times 10^7 \, \text{J}}{1.4 \times 10^8 \, \text{J/kg}} \approx 0.13 \, \text{kg} $。
(2) $ v = \frac{s}{t} = \frac{6 \times 10^3 \, \text{m}}{5 \times 60 \, \text{s}} = 20 \, \text{m/s} $，$ F = \frac{P}{v} = \frac{36 \times 10^3 \, \text{W}}{20 \, \text{m/s}} = 1800 \, \text{N} $
(3) $ W = Pt = 36 \times 10^3 \, \text{W} \times 5 \times 60 \, \text{s} = 1.08 \times 10^7 \, \text{J} $，$ Q_{\text{放}} = \frac{W}{\eta} = \frac{1.08 \times 10^7 \, \text{J}}{60\%} = 1.8 \times 10^7 \, \text{J} $，$ m_{\text{氢}} = \frac{Q_{\text{放}}}{q_{\text{氢}}} = \frac{1.8 \times 10^7 \, \text{J}}{1.4 \times 10^8 \, \text{J/kg}} \approx 0.13 \, \text{kg} $
解析：(1) 该车静止在水平地面上时对地面的压强 $ p = \frac{F}{S} = \frac{G}{S} = \frac{mg}{S} = \frac{1.8 \times 10^3 \, \text{kg} \times 10 \, \text{N/kg}}{0.1 \, \text{m}^2} = 1.8 \times 10^5 \, \text{Pa} $。(2) 本次测试中，该车匀速直线行驶 $ 6 \, \text{km} $ 用时 $ 5 \, \text{min} $，则该车行驶的速度 $ v = \frac{s}{t} = \frac{6 \times 10^3 \, \text{m}}{5 \times 60 \, \text{s}} = 20 \, \text{m/s} $；牵引力做功的功率恒为 $ 36 \, \text{kW} $，则该车受到的牵引力 $ F = \frac{P}{v} = \frac{36 \times 10^3 \, \text{W}}{20 \, \text{m/s}} = 1800 \, \text{N} $。(3) 本次测试中，牵引力做的功 $ W = Pt = 36 \times 10^3 \, \text{W} \times 5 \times 60 \, \text{s} = 1.08 \times 10^7 \, \text{J} $，氢燃料完全燃烧放出的热量 $ Q_{\text{放}} = \frac{W}{\eta} = \frac{1.08 \times 10^7 \, \text{J}}{60\%} = 1.8 \times 10^7 \, \text{J} $，消耗氢燃料的质量 $ m_{\text{氢}} = \frac{Q_{\text{放}}}{q_{\text{氢}}} = \frac{1.8 \times 10^7 \, \text{J}}{1.4 \times 10^8 \, \text{J/kg}} \approx 0.13 \, \text{kg} $。
3. 国庆期间,小明和爸爸进行自驾游,汽车以120km/h的速度匀速直线行驶了2h,从油量指示表上得出消耗的汽油为16L.汽油密度ρ取$0.75×10^3kg/m^3,$汽油热值q取$4.5×10^7J/kg,$小明了解到发动机的效率为30%.
(1)求16L汽油完全燃烧放出的热量.
(2)求汽车发动机输出的有用功.
(3)求汽车在行驶过程中所受的阻力.
(1)求16L汽油完全燃烧放出的热量.
(2)求汽车发动机输出的有用功.
(3)求汽车在行驶过程中所受的阻力.
答案
(1) $ m = \rho V = 0.75 \times 10^3 \, \text{kg/m}^3 \times 16 \times 10^{-3} \, \text{m}^3 = 12 \, \text{kg} $，$ Q_{\text{放}} = mq = 12 \, \text{kg} \times 4.5 \times 10^7 \, \text{J/kg} = 5.4 \times 10^8 \, \text{J} $
(2) $ W_{\text{有用}} = \eta Q_{\text{放}} = 30\% \times 5.4 \times 10^8 \, \text{J} = 1.62 \times 10^8 \, \text{J} $
(3) $ s = vt = 120 \, \text{km/h} \times 2 \, \text{h} = 240 \, \text{km} = 2.4 \times 10^5 \, \text{m} $，$ F = \frac{W_{\text{有用}}}{s} = \frac{1.62 \times 10^8 \, \text{J}}{2.4 \times 10^5 \, \text{m}} = 675 \, \text{N} $，$ f = F = 675 \, \text{N} $
解析：(1) 汽油质量 $ m = \rho V = 0.75 \times 10^3 \, \text{kg/m}^3 \times 16 \times 10^{-3} \, \text{m}^3 = 12 \, \text{kg} $，汽油完全燃烧释放的热量 $ Q_{\text{放}} = mq = 12 \, \text{kg} \times 4.5 \times 10^7 \, \text{J/kg} = 5.4 \times 10^8 \, \text{J} $。(2) 发动机的效率为 $ 30\% $，发动机输出的有用功 $ W_{\text{有用}} = \eta Q_{\text{放}} = 30\% \times 5.4 \times 10^8 \, \text{J} = 1.62 \times 10^8 \, \text{J} $。(3) 汽车行驶的路程 $ s = vt = 120 \, \text{km/h} \times 2 \, \text{h} = 240 \, \text{km} = 2.4 \times 10^5 \, \text{m} $，由 $ W = Fs $ 可知，汽车所受的牵引力 $ F = \frac{W_{\text{有用}}}{s} = \frac{1.62 \times 10^8 \, \text{J}}{2.4 \times 10^5 \, \text{m}} = 675 \, \text{N} $，因汽车匀速直线行驶，故汽车处于平衡状态，所受的阻力 $ f = F = 675 \, \text{N} $。
(2) $ W_{\text{有用}} = \eta Q_{\text{放}} = 30\% \times 5.4 \times 10^8 \, \text{J} = 1.62 \times 10^8 \, \text{J} $
(3) $ s = vt = 120 \, \text{km/h} \times 2 \, \text{h} = 240 \, \text{km} = 2.4 \times 10^5 \, \text{m} $，$ F = \frac{W_{\text{有用}}}{s} = \frac{1.62 \times 10^8 \, \text{J}}{2.4 \times 10^5 \, \text{m}} = 675 \, \text{N} $，$ f = F = 675 \, \text{N} $
解析：(1) 汽油质量 $ m = \rho V = 0.75 \times 10^3 \, \text{kg/m}^3 \times 16 \times 10^{-3} \, \text{m}^3 = 12 \, \text{kg} $，汽油完全燃烧释放的热量 $ Q_{\text{放}} = mq = 12 \, \text{kg} \times 4.5 \times 10^7 \, \text{J/kg} = 5.4 \times 10^8 \, \text{J} $。(2) 发动机的效率为 $ 30\% $，发动机输出的有用功 $ W_{\text{有用}} = \eta Q_{\text{放}} = 30\% \times 5.4 \times 10^8 \, \text{J} = 1.62 \times 10^8 \, \text{J} $。(3) 汽车行驶的路程 $ s = vt = 120 \, \text{km/h} \times 2 \, \text{h} = 240 \, \text{km} = 2.4 \times 10^5 \, \text{m} $，由 $ W = Fs $ 可知，汽车所受的牵引力 $ F = \frac{W_{\text{有用}}}{s} = \frac{1.62 \times 10^8 \, \text{J}}{2.4 \times 10^5 \, \text{m}} = 675 \, \text{N} $，因汽车匀速直线行驶，故汽车处于平衡状态，所受的阻力 $ f = F = 675 \, \text{N} $。
