Question

3. Ледяная горка составляет с горизонтом угол 10°. По горке пускают вверх камень, который, поднявшись на некоторую высоту, соскальзывает по тому же пути вниз. Определить коэффициент трения, если время спуска в 2 раза больше времени подъема. Если решите то будет шикарно. Зарание благодарю )

Answer 1

S=(a1²t1²)/2=(a2²t2²)⇒a1/a2=(t2/t1)²=(2/1)²=4

для подъёма ma1=kmgcosα+mgsinα

для спуска ma2=mgsinα-kmgcosα
делим 1 на 2 ⇒ kcosα+sinα=4sinα-4kcos⇒k=0,6tgα

Answer 2

Problem Analysis

We are given that a stone is launched up a hill with an angle of 10° with the horizontal. The stone then slides back down the same path. We need to determine the coefficient of friction if the time taken to slide down is twice the time taken to go up.

Solution

To solve this problem, we can use the concept of conservation of mechanical energy. The total mechanical energy of the stone at the top of the hill is equal to the total mechanical energy at the bottom of the hill.

The total mechanical energy of the stone can be expressed as the sum of its potential energy and kinetic energy:

E = PE + KE

At the top of the hill, all the stone's energy is in the form of potential energy, and at the bottom of the hill, all the energy is in the form of kinetic energy.

Let's denote the initial velocity of the stone as v and the height of the hill as h.

At the top of the hill: - Potential energy (PE) = mgh (where m is the mass of the stone and g is the acceleration due to gravity) - Kinetic energy (KE) = 0 (the stone is momentarily at rest)

At the bottom of the hill: - Potential energy (PE) = 0 (the stone is at the lowest point, so its potential energy is zero) - Kinetic energy (KE) = 1/2 * mv^2 (where v is the final velocity of the stone)

Since the stone slides back down the same path, the final velocity at the bottom of the hill is the same as the initial velocity at the top of the hill.

Using the conservation of mechanical energy, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill:

mgh = 1/2 * mv^2

Simplifying the equation, we find:

v^2 = 2gh

Now, let's consider the time taken to go up the hill and slide down the hill. We are given that the time taken to slide down is twice the time taken to go up. Let's denote the time taken to go up as t.

The distance traveled up the hill is equal to the distance traveled down the hill, so we can write:

d = d

Using the equations of motion, we can express the distance traveled as a function of time and initial velocity:

d = vt - 1/2 * gt^2

For the stone going up the hill, the initial velocity is v and the acceleration due to gravity is -g (negative because it acts in the opposite direction).

For the stone sliding down the hill, the initial velocity is 0 and the acceleration due to gravity is g.

Using the given information that the time taken to slide down is twice the time taken to go up, we can write:

2t = t

Simplifying the equation, we find:

t = 2t

Now, let's substitute the values of v^2 and t into the equation for distance traveled:

d = vt - 1/2 * gt^2

d = v(2t) - 1/2 * g(2t)^2

d = 2vt - 2gt^2

Since the distance traveled up the hill is equal to the distance traveled down the hill, we can write:

2vt - 2gt^2 = vt - 1/2 * gt^2

Simplifying the equation, we find:

v = 3gt

Now, we have two equations:

v^2 = 2gh

v = 3gt

We can solve these equations simultaneously to find the value of g.

Calculation

Let's solve the equations to find the value of g.

Substituting v = 3gt into v^2 = 2gh, we get:

(3gt)^2 = 2gh

Simplifying the equation, we find:

9g^2t^2 = 2gh

Dividing both sides of the equation by t^2, we get:

9g^2 = 2gh / t^2

Simplifying further, we find:

g = 2h / (9t^2)

Now, we can substitute the given values of h and t to find the value of g.

Answer

The coefficient of friction can be determined using the equation g = 2h / (9t^2), where h is the height of the hill and t is the time taken to go up the hill.

Please provide the values of h and t so that we can calculate the coefficient of friction.