Question

Брусок,находящийся на наклонной плоскости с углом наклона 30 градусов и коэффициентом трения 0,2 начал движение вниз из состояния покоя.Какое расстояние вдоль наклонной плоскости пройдет брусок к тому моменту,когда его скорость станет равной 5м.с

Answer 1

A=3.26 м/c^2. S=3.8(округлим 4м). 

Answer 2

Calculation of the Distance Traveled by the Block

To calculate the distance traveled by the block along the inclined plane, we need to consider the initial conditions and the motion of the block.

Given: - Angle of inclination of the plane: 30 degrees - Coefficient of friction: 0.2 - Initial velocity of the block: 0 m/s - Final velocity of the block: 5 m/s

We can use the equations of motion to solve for the distance traveled by the block.

1. First, let's find the acceleration of the block along the inclined plane. The acceleration can be calculated using the formula:

a = g * sin(θ) - μ * g * cos(θ)

where: - a is the acceleration - g is the acceleration due to gravity (approximately 9.8 m/s^2) - θ is the angle of inclination (30 degrees) - μ is the coefficient of friction (0.2)

Plugging in the values, we get:

a = (9.8 m/s^2) * sin(30) - (0.2) * (9.8 m/s^2) * cos(30)

Simplifying the equation, we find:

a ≈ 4.9 m/s^2

2. Next, we can use the equation of motion to find the distance traveled by the block. The equation is:

v^2 = u^2 + 2as

where: - v is the final velocity (5 m/s) - u is the initial velocity (0 m/s) - a is the acceleration (4.9 m/s^2) - s is the distance traveled

Plugging in the values, we get:

(5 m/s)^2 = (0 m/s)^2 + 2 * (4.9 m/s^2) * s

Simplifying the equation, we find:

25 m^2/s^2 = 9.8 m/s^2 * s

Solving for s, we get:

s ≈ 2.55 m

Therefore, the block will travel approximately 2.55 meters along the inclined plane before its velocity becomes 5 m/s.