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USAPhO 1999, Полуфинал — Задача 1. A1

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1999 Semi-Final Exam

Условие

A small crate of weight 5.0 N is released from rest at a height 2.0 m up a smooth inclined plane. The crate slides down the plane and across a smooth 1.50-m floor to a rough plane where a spring is located. The bottom of the spring is located at a height of 0.50 m. The spring constant is 20 N/m. The coefficient of kinetic friction between the rough plane and crate is $1/\sqrt{3}$ and the coefficient of static friction is $1/\sqrt{2}$ .

a. What is the maximum height the crate reaches on the right-hand inclined plane? [10 т.]

b. Where does the crate permanently come to rest? How many times does the crate travel up the rough plane? [10 т.]

(contributed by Mary Mogge)

Отговори

Решение

a. Let $Mg = 5.0\ \mathrm{N}$ = weight of the crate

$h = 2.0\ \mathrm{m}$ = the initial height of the crate

$h'$ = maximum height the crate reaches on the right hand side

$s$ = distance traveled along the rough plane to highest point.

$x$ = maximum compression of the spring

h' = s\sin 30^{\circ} = s\left(\tfrac{1}{2}\right) \qquad s = 2h'

x = s - \left(\frac{0.5m}{\sin 30^{\circ}}\right) = s - 1m = 2h' - 1m

To find the final height h', apply the work energy theorem

U_G' + U_{sp}' = U_G + W_f

Mgh' + \tfrac{1}{2}kx^2 = Mgh - \mu_k F_N s

F_N = Mg\cos 30^{\circ} = Mg\frac{\sqrt{3}}{2}

Mgh' + \tfrac{1}{2}k\left(2h' - 1\right)^2 = Mgh - \left(\frac{1}{\sqrt{3}}\right)\left(Mg\frac{\sqrt{3}}{2}\right)\left(2h'\right)

Mgh' + \tfrac{1}{2}k\left(2h' - 1\right)^2 = Mgh - Mgh'

2Mgh' + \tfrac{1}{2}k\left(2h' - 1\right)^2 = Mgh

2(5N)h' + \tfrac{1}{2}\left(20N/m\right)\left(2h' - 1\right)^2 = \left(5N\right)\left(2m\right)

h' + \left(2h' - 1\right)^2 = 1 \qquad \text{where } h' \text{ is in meters}

4h'^2 - 3h' = 0

Since $h'$ can't be zero

h' = \left(\tfrac{3}{4}\right)m

b. Initially it has $Mgh = (5N)(2m) = 10J$ of total mechanical energy. Sliding up to the highest point on the right hand side, the crate loses total mechanical energy

\mu_k F_N s = \mu_k\left(Mg\cos\theta\right)\left(2h'\right) = \left(1/\sqrt{3}\right)Mg\left(\sqrt{3}/2\right)\left(2h'\right)

= Mgh' = (5N)(0.75m) = 3.75J

(Note: On the first trip up the plane, the crate reached the spring which provided an additional force of kx down the plane. At maximum extension this force is 10 N, sufficient to overcome static friction.)

Sliding back down it loses another 3.75 J of total mechanical energy. Therefore it has

10J - 2\left(3.75J\right) = 2.5J

at the bottom of the right hand plane. The total mechanical energy remains unchanged until the crate returns to the right hand plane and begins to slide up it. Using our prior result that as the crate slides up the plane

\mu_k F_N s = Mgh'

That is, as the crate rises half the kinetic energy becomes potential and half is dissipated by friction, it will rise to a height given by

\left(\tfrac{1}{2}\right)\left(2.5J\right) = Mgh' = \left(5N\right)h'

h' = 0.25m

It stops before it reaches the spring. At this point, the only two forces acting on the crate are the gravitational force and the frictional force. The component of the weight parallel to the plane is

F_{G|} = Mg\sin 30^{\circ} = (5N)\left(\tfrac{1}{2}\right) = 2.5N

down the plane. the maximum force of static friction is

f_{s\max} = \mu_s Mg\cos 30^{\circ} = \left(\tfrac{1}{\sqrt{2}}\right)(5N)\left(\tfrac{\sqrt{3}}{2}\right) = 3.1N

F_{G|} < f_{s\max}

The crate comes to rest a height 0.25m up the right hand plane. It has slid down the smooth plane, up the rough plane, temporarily compressing the spring, back down the rough plane, across the horizontal surface, up and down the smooth plane, back across the horizontal surface, and up the rough plane a second time before it finally stops.

Автори

The 1999 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Dr. Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Dr. Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Dr. Warren Turner – Physics Teacher at St. Paul's School, Concord, NH. The coaches would also like to thank former academic director, Dr. Larry Kirkpatrick for his many helpful comments.

Оригинал в Архива: Semifinal 1999.pdf · официални решения: Semifinal 1999 Solutions.pdf