Question

Хоккеист ударяет по шайбе массой 200 г и сообщает ей скорость 10 м/с. В результате удара шайба проскользила 50 м пути и остановилась. Определи силу трения, действующую на шайбу, и коэффициент трения между льдом и шайбой. Ответ запиши в указанном месте. Ускорение свободного падения равно 10 м/с2.

Answer 1

Ответ:

ответ на фото

Объяснение:

Answer 2

Problem Analysis

We are given the mass of the hockey puck (200 g), its initial velocity (10 m/s), and the distance it slides (50 m) before coming to a stop. We need to determine the force of friction acting on the puck and the coefficient of friction between the ice and the puck.

Solution

To solve this problem, we can use the equations of motion and the concept of work done by friction.

1. We can start by calculating the initial kinetic energy of the puck using the formula: Kinetic energy = (1/2) * mass * velocity^2. Plugging in the values, we get: Kinetic energy = (1/2) * 0.2 kg * (10 m/s)^2.

2. Next, we can calculate the work done by friction using the formula: Work done by friction = force of friction * distance. Since the work done by friction is equal to the change in kinetic energy, we can equate the two equations: Force of friction * distance = (1/2) * mass * velocity^2.

3. Rearranging the equation, we can solve for the force of friction: Force of friction = (1/2) * mass * velocity^2 / distance.

4. Finally, we can calculate the coefficient of friction using the formula: Coefficient of friction = force of friction / (mass * acceleration due to gravity). Since the acceleration due to gravity is given as 10 m/s^2, we can plug in the values to find the coefficient of friction.

Let's calculate the force of friction and the coefficient of friction using the given values.

Calculation

1. Calculating the initial kinetic energy: Kinetic energy = (1/2) * 0.2 kg * (10 m/s)^2.

2. Calculating the force of friction: Force of friction = (1/2) * 0.2 kg * (10 m/s)^2 / 50 m.

3. Calculating the coefficient of friction: Coefficient of friction = (1/2) * 0.2 kg * (10 m/s)^2 / (0.2 kg * 10 m/s^2).

Answer

After performing the calculations, we find that the force of friction acting on the puck is 0.2 N and the coefficient of friction between the ice and the puck is 0.5.

Please note that the coefficient of friction is dimensionless and represents the ratio of the force of friction to the normal force between the two surfaces.