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M2 2023 Oct Q5

A Level / Edexcel / M2

IAL 2023 Oct Paper · Question 5

题目

Problem

5.

A cyclist is travelling on a straight horizontal road and working at a constant rate of 500 W500\,\mathrm{W}.

The total mass of the cyclist and her cycle is 80 kg80\,\mathrm{kg}.

The total resistance to the motion of the cyclist is modelled as a constant force of magnitude 60 N60\,\mathrm{N}.

(a) Using this model, find the acceleration of the cyclist at the instant when her speed is 6 m s−16\,\mathrm{m\,s^{-1}}

(4)

On the following day, the cyclist travels up a straight road from a point AA to a point BB.

The distance from AA to BB is 20 km20\,\mathrm{km}.

Point AA is 500 m500\,\mathrm{m} above sea level and point BB is 800 m800\,\mathrm{m} above sea level.

The cyclist starts from rest at AA.

At the instant she reaches BB her speed is 8 m s−18\,\mathrm{m\,s^{-1}}

The total resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude 60 N60\,\mathrm{N}.

(b) Using this model, find the total work done by the cyclist in the journey from AA to BB.

(5)

Later on, the cyclist is travelling up a straight road which is inclined at an angle α\alpha to the horizontal, where sin⁡α=120\sin\alpha=\frac1{20}

The cyclist is now working at a constant rate of PP watts and has a constant speed of 7 m s−17\,\mathrm{m\,s^{-1}}

The total resistance to the motion of the cyclist from non-gravitational forces is again modelled as a constant force of magnitude 60 N60\,\mathrm{N}.

(c) Using this model, find the value of PP

(4)
题目中文翻译

一名骑车人在水平直路上行驶，以恒定功率 500 W500\,\mathrm{W} 做功。

骑车人与自行车的总质量为 80 kg80\,\mathrm{kg}。

将骑车人所受的总运动阻力视为大小为 60 N60\,\mathrm{N} 的恒力。

(a) 利用此模型，求她的速率为 6 m s−16\,\mathrm{m\,s^{-1}} 的瞬间的加速度。

(4)

第二天，骑车人沿直路从 AA 点上坡至 BB 点。

AA 到 BB 的距离为 20 km20\,\mathrm{km}。

AA 点海拔 500 m500\,\mathrm{m}，BB 点海拔 800 m800\,\mathrm{m}。

骑车人从 AA 点由静止出发。

她到达 BB 的瞬间的速率为 8 m s−18\,\mathrm{m\,s^{-1}}。

将骑车人所受的非重力总运动阻力视为大小为 60 N60\,\mathrm{N} 的恒力。

(b) 利用此模型，求骑车人从 AA 到 BB 的过程中所做的总功。

(5)

后来，骑车人沿一条与水平面成 α\alpha 角的直路上坡，其中 sin⁡α=120\sin\alpha=\frac1{20}。

此时骑车人以恒定功率 PP 瓦做功，并以恒定速率 7 m s−17\,\mathrm{m\,s^{-1}} 行驶。

仍将骑车人所受的非重力总运动阻力视为大小为 60 N60\,\mathrm{N} 的恒力。

(c) 利用此模型，求 PP 的值。

(4)

解答

(a)

解法一

思路

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恒功率不代表恒牵引力或恒加速度。先求速率为 66 时的瞬时牵引力，再用牛顿第二定律。

答题过程

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Let the driving force at speed 6 m s−1^{-1} be FF N. From power =Fv=Fv,

F=5006 N.F=\frac{500}{6}\text{ N}.

Resolving along the horizontal road,

F−60=80a.F-60=80a.

Therefore

a=500/6−6080,a=\frac{500/6-60}{80},

so

a=724 m s−2.\boxed{a=\frac7{24}\text{ m s}^{-2}}.

(b)

解法一

思路

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骑车人做的功同时用于增加动能、增加重力势能和克服阻力。路程 2020 千米用于阻力功，升高 300300 米用于势能，不能混用。

答题过程

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The distance travelled is 20000 m, and the vertical rise is

800−500=300 m.800-500=300\text{ m}.

The gain in kinetic energy is

ΔK=12(80)(82)−0=2560 J.\Delta K=\frac12(80)(8^2)-0=2560\text{ J}.

The gain in gravitational potential energy is

ΔU=80g(300)=235200 J.\Delta U=80g(300)=235200\text{ J}.

The work done against resistance is

WR=60(20000)=1200000 J.W_R=60(20000)=1200000\text{ J}.

By the work-energy principle, the cyclist’s total work is

W=ΔK+ΔU+WR.W=\Delta K+\Delta U+W_R.

Thus

W=2560+235200+1200000=1437760 J.W=2560+235200+1200000=1437760\text{ J}.

To 3 significant figures,

W=1.44×106 J.\boxed{W=1.44\times10^6\text{ J}}.

(c)

解法一

思路

展开

本问速率恒定，沿坡合力为零。牵引力须同时平衡非重力阻力和重力沿坡分量，再乘速率求功率。

答题过程

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At constant speed the acceleration is zero. Resolving up the road,

F−60−80gsin⁡α=0.F-60-80g\sin\alpha=0.

Using F=P/7F=P/7 and sin⁡α=1/20\sin\alpha=1/20,

P7=60+80(9.8)(120)=99.2.\frac P7=60+80(9.8)\left(\frac1{20}\right)=99.2.

Hence P=694.4P=694.4 W, giving

P=694 W.\boxed{P=694\text{ W}}.