Question

Брусок соскальзывает с наклонной плоскости, угол наклона которой равен 45 °, ускорением 3,65 м / с2. Определите ко- ефициент трения бруска о плоскость.

Answer 1

По второму закону Ньютона (в векторной форме):
по оси OX: mg+Fтр=ma
по оси OY: N+Fтр=0
перепишем в проекциях
mg*sin(L)-Fтр=ma (1)
N-mg*cos(L)=0 (2)
из уравнения 2: N=mg*cos(L)
Fтр= k*N
Подставим в уравнение (1): mg*sin(L)-ma=k*mg*cos(L)
k=(g*sin(L) - a)/g*cos(L)= (т.к. L=45 градусов, то sin=cos) = 1 - a/(g*sin(L))= = 1- 3,65/9,8*(sqrt2/2)=1-7,3/13,859=1-0,527=0,473

Answer 2

Calculation of the Coefficient of Friction

To determine the coefficient of friction between the block and the inclined plane, we can use the given information about the angle of inclination and the acceleration of the block.

The formula for the acceleration of an object on an inclined plane is:

a = g * sin(θ) - μ * g * cos(θ)

Where: - a is the acceleration of the block - g is the acceleration due to gravity (approximately 9.8 m/s^2) - θ is the angle of inclination (45° in this case) - μ is the coefficient of friction

We are given that the acceleration of the block is 3.65 m/s^2. Plugging in the values into the formula, we can solve for the coefficient of friction:

3.65 = 9.8 * sin(45°) - μ * 9.8 * cos(45°)

Simplifying the equation:

3.65 = 9.8 * 0.707 - μ * 9.8 * 0.707

3.65 = 6.92 - μ * 6.92

Rearranging the equation to solve for μ:

μ * 6.92 = 6.92 - 3.65

μ = (6.92 - 3.65) / 6.92

μ ≈ 0.472

Therefore, the coefficient of friction between the block and the inclined plane is approximately 0.472.

Please note that the sources provided did not contain specific information related to this calculation. The calculation is based on the principles of physics and the formula for the acceleration of an object on an inclined plane.