Sliding Block on an Incline Problem
What is the magnitude of the friction force along the plane when a block slides down an incline?
A block with a mass of 5.0 kg slides down a 37° incline with an acceleration of 5.6 m/s². The coefficient of kinetic friction between the block and the inclined surface is 0.050. What is the magnitude of the friction force along the plane?
Answer:
The magnitude of the friction force along the plane is 2 N.
When a block of 5.0 kg slides down a 37° incline, we can analyze the forces acting on the block to determine the friction force along the plane. The acceleration of the block is 5.6 m/s², with sin 37° = 0.6 and cos 37° = 0.8.
The equation for the forces along the direction perpendicular to the plane is:
R - mg cos θ = 0
Where: R is the reaction force, mg cos θ is the component of the weight perpendicular to the plane, m = 5.0 kg (mass of the block), g = 9.8 m/s² (acceleration of gravity), θ = 37°.
Solving for R, we get:
R = mgcos θ = (5.0)(9.8)(0.8) = 39.2 N
The frictional force is given by:
F = μR
Where μ = 0.050 (coefficient of friction).
Substituting the values, we find:
F = (0.050)(39.2) = 2 N
Therefore, the magnitude of the friction force along the plane when a block slides down the incline is 2 N.