Answer:
Here's a free body diagram for the block:
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Step-by-step explanation:
The forces acting on the block are:
Weight (W): this is the force due to gravity acting on the block and is equal to m * g, where m is the mass of the block (5.0 kg) and g is the acceleration due to gravity (approximately 9.81 m/s²). In the diagram, this force acts straight down.
Normal force (N): this is the force exerted by the incline on the block perpendicular to the surface of the incline. In the diagram, this force acts perpendicular to the incline and away from it.
Force of static friction (f_s): this is the force that opposes the motion of the block and acts parallel to the surface of the incline. Its magnitude is given by f_s ≤ μ_s * N, where μ_s is the coefficient of static friction. In the diagram, this force acts up the incline.
The net force acting on the block is the vector sum of all forces acting on it. In this case, we can resolve the forces into components parallel and perpendicular to the incline:
Parallel to incline: f_s = f_parallel
Perpendicular to incline: N - W*cosθ = f_perpendicular
where θ is the angle of the incline (30°), and f_parallel and f_perpendicular are the parallel and perpendicular components of the force of static friction, respectively.
To determine whether the block will slide or remain at rest, we need to compare the magnitude of the force of static friction with the maximum possible force of static friction, which is μ_s * N = 0.5 * (m * g * cosθ + N * sinθ). If f_s ≤ μ_s * N, then the block will remain at rest. If f_s > μ_s * N, then the block will slide.
Since the block is at rest, we know that f_s = f_parallel = 0. Therefore, N = W*cosθ, and the force of static friction is at its maximum possible value, which is:
f_s = μ_s * N = 0.5 * (m * g * cosθ + N * sinθ) = 0.5 * (5.0 kg * 9.81 m/s² * cos(30°) + W * sin(30°)) ≈ 23.92 N
Since f_s is less than the maximum possible force of static friction, the block will remain at rest and not slide down the incline.