A 1.8-kg block is released from rest at the top of a rough 30$\degree$ 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

We can use the formula for force of friction: 
f_friction = μ_friction * N

where μ_friction is the coefficient of friction and N is the normal force. 
The normal force can be found using the component of gravity that is perpendicular to the incline:

N = m * g * cos(theta)

where m is the mass, g is the acceleration due to gravity, and theta is the angle of the incline.

Plugging in the given values, we get: 
N = (1.8 kg)(9.81 m/s^2) * cos(30°) 
N = 15.68 N

Now we can use the fact that the acceleration down the incline is 3.0 m/s^2:

f_net = m * a 
f_net = (1.8 kg) * (3.0 m/s^2) 
f_net = 5.4 N

The net force is the force of gravity minus the force of friction:

f_net = f_gravity - f_friction

Plugging in the known values, we obtain:

5.4 N = (1.8 kg)(9.81 m/s^2) * sin(30°) - f_friction

Solving for f_friction, we get:

f_friction = 3.39 N

Rounding this to one significant figure gives us the final answer of:

f_friction = 3.4 N

Therefore, the best choice is (C).

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