The coefficient of static friction between a brick and a wooden board is 0.40 and the coefficient of kinetic friction between the brick and board is $0.30 .$ You place the brick on the board and slowly lift one end of the board off the ground until the brick starts to slide down the board. (a) What angle does the board make with the ground when the brick starts to slide? (b) What is the acceleration of the brick as it slides down the board?

To find the angle at which the brick starts to slide down the board, we can use the coefficient of static friction. The maximum force of static friction is given by: \(F_{s(max)} = μ_{s} mg\) where \(F_{s(max)}\) is the maximum force of static friction, \(μ_{s}\) is the coefficient of static friction, \(m\) is the mass of the brick, and \(g\) is the acceleration due to gravity (approximately 9.81 m/s²). At the point when the brick starts to slide, the force of static friction equals the force of gravity trying to pull the brick down the inclined plane. The force of gravity acting along the inclined plane is given by: \(F_{g} = mg\sin(\theta)\) where \(θ\) is the angle of inclination. So, we can set up the following equation and solve for the angle: \(μ_{s} mg = mg\sin(\theta)\)

Next, we will solve for the angle \(θ\). We can see that the mass \(m\) and gravitational constant \(g\) can be canceled from both sides of the equation: \(μ_{s} = \sin(\theta)\) Now, we can find the angle using the inverse sine function: \(θ = \sin^{-1}(μ_{s})\) Plugging in the given value of the coefficient of static friction: \(θ = \sin^{-1}(0.40)\) Calculating the angle: \(θ \approx 23.6°\)

To determine the acceleration, we can use the coefficient of kinetic friction, \(μ_{k}\). The kinetic friction force acting on the brick as it slides down the inclined plane is given by: \(F_{k} = μ_{k} mg\cos(\theta)\) The net force acting on the brick down the incline is the difference between gravitational force along the incline and the kinetic friction force, which can be written as: \(F_{net} = mg\sin(\theta) - μ_{k} mg\cos(\theta)\) Since \(F = ma\), the acceleration down the incline is given by: \(a = \frac{F_{net}}{m}\) Plugging in the values for the forces: \(a = \frac{mg\sin(\theta) - μ_{k} mg\cos(\theta)}{m}\) We can now plug in the values for \(μ_{k}\), \(g\), and \(θ\): \(a = \frac{9.81\sin(23.6°) - 0.30 \cdot 9.81\cos(23.6°)}{m}\) We notice that once again, the mass of the brick, \(m\), cancels out: \(a \approx 9.81\sin(23.6°) - 0.30 \cdot 9.81\cos(23.6°)\) Calculating the acceleration: \(a \approx 3.82\:m/s^2\)

Finally, we can write the answers to both parts of the question: (a) The angle at which the brick starts to slide is approximately \(23.6°\). (b) The acceleration of the brick as it slides down the board is approximately \(3.82\:m/s^2\).