This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Welcome back CALM🙂↔️ — been a while! Let's pick up where you left off. To find the acceleration of the brick, we need to analyze the forces acting on it along the inclined plane. We'll use the given mass, angle of inclination, and coefficient of friction. We will assume the acceleration due to gravity g = 9.8 \, m/s^2. Step 1: Determine and . Given = (1)/(3). We can construct a right-angled triangle where the opposite side is 1 and the adjacent side is 3. The hypotenuse h is calculated using the Pythagorean theorem: h = sqrt(1^2 + 3^2) = sqrt(1 + 9) = sqrt(10) Now we can find and : = oppositehypotenuse = (1)/(sqrt(10)) = adjacenthypotenuse = (3)/(sqrt(10)) Step 2: Calculate the normal force (N). The forces perpendicular to the inclined plane are the normal force (N) acting upwards and the component of gravity (mg ) acting downwards into the plane. Since there is no acceleration perpendicular to the plane, these forces balance: N - mg = 0 N = mg Step 3: Calculate the maximum static frictional force (f_s,max). The maximum static frictional force is given by f_s,max = N, where is the coefficient of friction. f_s,max = (mg ) Substitute the given value of = (1)/(3): f_s,max = (1)/(3) mg ( (3)/(sqrt(10)) ) = (mg)/(sqrt(10)) Step 4: Determine the component of gravity parallel to the plane (F_down). The component of gravity pulling the brick down the incline is mg . F_down = mg = mg ( (1)/(sqrt(10)) ) = (mg)/(sqrt(10)) Step 5: Compare forces and find acceleration. For the brick to move, the downward component of gravity must be greater than the maximum static frictional force (F_down > f_s,max). In this case, we found: F_down = (mg)/(sqrt(10)) f_s,max = (mg)/(sqrt(10)) Since F_down = f_s,max, the component of gravity pulling the brick down the incline is exactly equal to the maximum static frictional force. This means the net force along the incline is zero. As the brick is released from rest, it will not start to move. Therefore, its acceleration is zero. Applying Newton's second law along the incline: F_ = ma F_down - f_s,max = ma (mg)/(sqrt(10)) - (mg)/(sqrt(10)) = ma 0 = ma Since the mass m is not zero, the acceleration a must be zero. a = 0 \, m/s^2 The acceleration of the brick is 0 m/s^2. Send me the next one 📸