题目
Problem
A car of mass 1800 k g 1800\,\mathrm{kg} 1800 kg 300 k g 300\,\mathrm{kg} 300 kg α \alpha α sin α = 0.05 \sin\alpha=0.05 sin α = 0.05 800 N 800\,\mathrm{N} 800 N F N F\,\mathrm{N} F N 3000 N 3000\,\mathrm{N} 3000 N
(a) It is given that F = 100 F=100 F = 100
Find the acceleration of the car and the tension in the tow-bar.
(5)
(b) It is given instead that the total work done against F F F 50 m 50\,\mathrm{m} 50 m 6000 J 6000\,\mathrm{J} 6000 J 50 m 50\,\mathrm{m} 50 m 20 m s − 1 20\,\mathrm{m\,s^{-1}} 20 m s − 1
Use an energy method to find the speed of the car at the end of the 50 m 50\,\mathrm{m} 50 m
(5)
题目中文翻译
一辆质量为 1800 k g 1800\,\mathrm{kg} 1800 kg 300 k g 300\,\mathrm{kg} 300 kg α \alpha α sin α = 0.05 \sin\alpha=0.05 sin α = 0.05 800 N 800\,\mathrm{N} 800 N F N F\,\mathrm{N} F N 3000 N 3000\,\mathrm{N} 3000 N
(a) 已知 F = 100 F=100 F = 100
求汽车的加速度和牵引杆中的张力。
(5)
(b) 改为已知沿道路向上运动 50 m 50\,\mathrm{m} 50 m F F F 6000 J 6000\,\mathrm{J} 6000 J 50 m 50\,\mathrm{m} 50 m 20 m s − 1 20\,\mathrm{m\,s^{-1}} 20 m s − 1
使用能量方法，求汽车在这段 50 m 50\,\mathrm{m} 50 m
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
Where a numerical value for the acceleration due to gravity (g g g 10 m s − 2 10\,\mathrm{m\,s^{-2}} 10 m s − 2
需要使用重力加速度 g g g 10 m s − 2 10\,\mathrm{m\,s^{-2}} 10 m s − 2
解答
(a)
解法一
思路
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上坡时两个阻力与两个重力分量均向下。整体求加速度，再对拖车求张力。
答题过程
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For the whole system, resolving up the road gives
3000 − 800 − 100 − 2100 ( 10 ) ( 0.05 ) = 2100 a . \begin{align*}
&\,3000-800-100-2100(10)(0.05)\\[4mm]
=&\,2100a.
\end{align*} = 3000 − 800 − 100 − 2100 ( 10 ) ( 0.05 ) 2100 a . Hence
a = 1050 2100 = 0.5 m s − 2 . \boxed{a=\frac{1050}{2100}=0.5\,\mathrm{m\,s^{-2}}}. a = 2100 1050 = 0.5 m s − 2 . For the trailer,
T − 100 − 300 ( 10 ) ( 0.05 ) = 300 ( 0.5 ) , T-100-300(10)(0.05)=300(0.5), T − 100 − 300 ( 10 ) ( 0.05 ) = 300 ( 0.5 ) , so
T = 400 N . \boxed{T=400\,\mathrm{N}}. T = 400 N .
(b)
解法一
思路
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这是另一种阻力状态，不能沿用(a)的张力或加速度。按要求对整体写能量方程，直接用给定阻力总功。
答题过程
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The work done by the driving force is 3000 ( 50 ) = 150000 J 3000(50)=150000\,\mathrm{J} 3000 ( 50 ) = 150000 J 800 ( 50 ) = 40000 J 800(50)=40000\,\mathrm{J} 800 ( 50 ) = 40000 J
2100 ( 10 ) ( 50 ) ( 0.05 ) = 52500 J . 2100(10)(50)(0.05)=52500\,\mathrm{J}. 2100 ( 10 ) ( 50 ) ( 0.05 ) = 52500 J . The work-energy equation for the car and trailer is
1 2 ( 2100 ) ( v 2 − 20 2 ) = 150000 − 40000 − 6000 − 52500 = 51500. \begin{align*}
\frac12(2100)(v^2-20^2)
=&\,150000-40000\\[4mm]
&\,\hspace{2pt}-6000-52500\\[4mm]
=&\,51500.
\end{align*} 2 1 ( 2100 ) ( v 2 − 2 0 2 ) = = 150000 − 40000 − 6000 − 52500 51500. Thus
1050 v 2 = 471500. 1050v^2=471500. 1050 v 2 = 471500. v = 471500 1050 ≈ 21.2 m s − 1 . v=\sqrt{\frac{471500}{1050}}
\approx\boxed{21.2\,\mathrm{m\,s^{-1}}}. v = 1050 471500 ≈ 21.2 m s − 1 .
解法二
思路
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也可只对汽车写能量方程，但必须先求本状态的新张力，不能沿用(a)的数值。张力对汽车做负功。
答题过程
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The trailer’s resistance in this case is
F = 6000 50 = 120 N . F=\frac{6000}{50}=120\,\mathrm{N}. F = 50 6000 = 120 N . Let the common acceleration be a a a T T T
3000 − 900 − 800 − T = 1800 a , T − 150 − 120 = 300 a . \begin{align*}
3000-900-800-T=&\,1800a,\\[4mm]
T-150-120=&\,300a.
\end{align*} 3000 − 900 − 800 − T = T − 150 − 120 = 1800 a , 300 a . Adding the equations,
1030 = 2100 a , a = 103 210 . 1030=2100a,\qquad a=\frac{103}{210}. 1030 = 2100 a , a = 210 103 . Hence
T = 270 + 300 ( 103 210 ) = 2920 7 N . T=270+300\bigg(\frac{103}{210}\bigg)
=\frac{2920}{7}\,\mathrm{N}. T = 270 + 300 ( 210 103 ) = 7 2920 N . For the car alone, driving work less work against resistance, tension and gravity equals the increase in kinetic energy:
1 2 ( 1800 ) ( v 2 − 20 2 ) = 3000 ( 50 ) − 800 ( 50 ) − 2920 7 ( 50 ) − 1800 ( 10 ) ( 50 ) ( 0.05 ) = 309000 7 . \begin{align*}
\frac12(1800)(v^2-20^2)
=&\,3000(50)-800(50)\\[4mm]
&\,\hspace{2pt}-\frac{2920}{7}(50)\\[4mm]
&\,\hspace{4pt}-1800(10)(50)(0.05)\\[4mm]
=&\,\frac{309000}{7}.
\end{align*} 2 1 ( 1800 ) ( v 2 − 2 0 2 ) = = 3000 ( 50 ) − 800 ( 50 ) − 7 2920 ( 50 ) − 1800 ( 10 ) ( 50 ) ( 0.05 ) 7 309000 . Therefore
v 2 = 400 + 309000 7 ( 900 ) = 9430 21 . v^2=400+\frac{309000}{7(900)}=\frac{9430}{21}. v 2 = 400 + 7 ( 900 ) 309000 = 21 9430 . Taking the positive root,
v = 9430 21 ≈ 21.2 m s − 1 . v=\sqrt{\frac{9430}{21}}
\approx\boxed{21.2\,\mathrm{m\,s^{-1}}}. v = 21 9430 ≈ 21.2 m s − 1 . 