Съдържание

Условие Отговори Решение

USAPhO 2003, Четвъртфинал — Задача 1

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Всички материали от USAPhO 2003

AAPT AIP United States Physics Team 2003 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Moment of Inertia about Center of Mass – Uniform Object

(will not be provided on the second screening exam)

Disk	$\tfrac{1}{2} MR^2$
Sphere	$\tfrac{2}{5} MR^2$
Rod	$\tfrac{1}{12} ML^2$

Solutions to Problems

Unless explicitly told to use a specific method, any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 2.a., if a student wrote down $2v_o t - \frac{1}{2}gt^2 = H - v_o t - \frac{1}{2}gt^2$ , they should get 8 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A straight incline descending to the right at angle θ to a grey horizontal line; a small rectangular block sits at the upper-left end marked s = 0, and an arrow along the incline points down-right labelled s.

A small block of weight $mg$ is initially at rest at the starting point $s = 0$ shown above. It slides down an increasingly rough incline at an angle of $\theta$ to the horizontal. The coefficient of kinetic friction is

\mu = \alpha s,

where $\alpha$ is a constant and $s$ is measured down along the incline from the starting point $s = 0$ . The magnitude of the work done by a non-constant force can be found from the area under a force versus displacement graph.

a. Draw a diagram which shows all the forces acting on the block (free body diagram). Clearly label each force. Write down the magnitude of each force in terms of only the quantities given above. [9 т.]

b. Sketch a graph of the magnitude of the frictional force versus displacement. [3 т.]

c. Find the area under the graph and use it to determine the magnitude of the work done by the frictional force as a function of $s$ . [3 т.]

d. How far down the plane does the block slide before coming to rest? [10 т.]

Отговори

Решение

1. a.

Drawing and labeling the $mg$ vector in the downward direction or correctly drawing and labeling the components parallel to the plane $(mg\sin\theta)$ and perpendicular to the plane $(mg\cos\theta)$ . [3]

Drawing and labeling the normal force (any alternate symbol for $N$ may be used).

N = mg\cos\theta.

[3]

Drawing and labeling the friction force (any alternate symbol for $f$ may be used).

f = \mu mg\cos\theta

f = \alpha s mg\cos\theta.

[3]

b. Sketch of force versus displacement graph showing a linear function of $s$ . [3]

c. Finding the area under the graph by any method and equating it to $|W|$ . For example,

|W| = \mathit{Area} = \tfrac{1}{2}bh = \tfrac{1}{2}(s)(\alpha s mg\cos\theta)

|W| = \tfrac{1}{2}(\alpha mg\cos\theta)s^2.

[3]

(The absolute value signs around $W$ are not required.)

d. Using the work-energy theorem

\mathrm{PE_i + KE_i} + W = \mathrm{PE_f + KE_f}

[1]

“i” refers to the starting point and “f” refers to the stopping point, a distance $s$ down the plane. At both points the kinetic energy is zero

\mathrm{KE_i} = 0 \qquad \mathrm{KE_f} = 0.

[2]

Setting the potential energy at the stopping point equal to zero, the potential energies are

\mathrm{PE_i} = mgh \qquad \mathrm{PE_f} = 0.

[2]

where $h$ is the vertical displacement of the starting point above the stopping point

h = s\sin\theta.

[1]

The frictional force does negative work

W = -\tfrac{1}{2}(\alpha mg\cos\theta)s^2.

[1]

Combining terms:

mgs\sin\theta - \tfrac{1}{2}(\alpha mg\cos\theta)s^2 = 0

[1]

The $s = 0$ solution represents the starting point and can be eliminated.

Dividing by $mgs$ ,

\sin\theta - \tfrac{1}{2}(\alpha\cos\theta)s = 0

[1]

Therefore

s = \frac{2\sin\theta}{\alpha\cos\theta} = \frac{2}{\alpha}\tan\theta

[1]

Free body diagram: a vertical downward arrow labelled mg with angle θ marked, an arrow up and to the right labelled N = mgcosθ, and an arrow up-left along the incline labelled f = μmgcosθ.

Graph for part b: vertical axis f = αsmgcosθ, horizontal axis s; a straight line rising from the origin, the area under it shaded as a triangle.

Оригинал в Архива: Quarterfinal 2003.pdf · официални решения: Quarterfinal 2003 Solutions.pdf