A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s² down the incline. Determine the magnitude of the force of friction acting on the block. A) 4.2 N B) 3.0 N C) 3.4 N D) 3.8 N E) 2.3 N

Question

Quizplus · Accepted Answer

The force of gravity down the incline can be resolved into two components: a force parallel to the incline and a force perpendicular to the incline. Since the block is sliding down the incline, we know the force of friction acts up the incline to oppose the motion. 
Fnet = ma
The net force acting on the block is the force of gravity parallel to the incline minus the force of friction: 
Fnet = mgsin(30°) - f 
where m = 1.8 kg, g = 9.8 m/s^2, and the angle of the incline is 30°. 
Plugging in the given acceleration a = 3.0 m/s^2: 
3.0 m/s^2 = (9.8 m/s^2)sin(30°) - f/1.8 kg 
Solving for f: 
f = (9.8 m/s^2)sin(30°) - (3.0 m/s^2)(1.8 kg) 
f ≈ 3.4 N

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