题目
Problem
A van of mass 2500kg travelling at speed vms−1 experiences a resistance force of kv2N. The constant power of the van’s engine is 62.5kW.
(a) The steady speed that the van could maintain when moving along a straight horizontal road is 50ms−1.
Show that k=0.5, and find the acceleration of the van when its speed is 25ms−1 on this straight horizontal road.
(4)
The van begins to ascend a hill inclined at an angle θ∘ to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is 20ms−1, and its acceleration is 5ams−2. Later, on the same hill, the speed of the van is 30ms−1 and its acceleration is ams−2. The power of the van’s engine remains at 62.5kW, and the resistance force remains at 0.5v2N.
(b) Find the value of a and the value of θ.
(5)
题目中文翻译
一辆质量为 2500kg 的厢式货车以 vms−1 的速率行驶时,受到 kv2N 的阻力。货车发动机的功率恒为 62.5kW。
(a) 货车沿水平直路行驶时能够维持的稳定速率为 50ms−1。
证明 k=0.5,并求货车在这条水平直路上以 25ms−1 的速率行驶时的加速度。
(4)
货车开始沿一条与水平面成 θ∘ 角的山坡上行,运动方向沿山坡的一条最大坡度线。货车在坡底的速率为 20ms−1,加速度为 5ams−2。之后,在同一山坡上,货车的速率为 30ms−1,加速度为 ams−2。货车发动机的功率仍为 62.5kW,阻力仍为 0.5v2N。
(b) 求 a 和 θ 的值。
(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) is needed, use 10ms−2.
需要使用重力加速度 g 的数值时,取 10ms−2。
解答
(a)
解法一
思路
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稳速时牵引力等于k50²,由功率关系求k;到25时牵引力和阻力均需重新计算。
答题过程
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At steady speed 50ms−1, the driving force equals the resistance. Hence
62500=(k⋅502)(50),k=0.5,
as required.
At 25ms−1, Newton’s second law gives
2562500−0.5(25)2=2500a25.
Therefore
a25=25002500−312.5=0.875 m s−2.
(b)
解法一
思路
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同一坡的重力分量相同,列20与30速率的两条方程相减消去角度,再回代。
答题过程
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Resolving up the hill at the two instants gives
2062500−0.5(20)2−25000sinθ=12500a,(1)
3062500−0.5(30)2−25000sinθ=2500a.(2)
Subtracting (2) from (1),
10000a==(3125−200)−(3062500−450)33875.
Thus
a=24031=0.129166667…≈0.129.
Substituting into (2),
25000sinθ==3062500−450−2500240311215725.
Hence
θ=3.004630723…∘≈3.0∘ (1 d.p.).