Slanted surfaces with friction: A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic friction is 0.19, at what rate does the box accelerate down the slope? A) 4.7 m/s² B) 5.2 m/s² C) 5.5 m/s² D) 6.2 m/s²

Question

Quizplus · Accepted Answer

The gravitational force acting on the box is given by:

F_gravity = m * g * sin(theta)
where m = 6.0 kg is the mass of the box, g = 9.81 m/s^2 is the acceleration due to gravity, and theta = 39° is the angle of the inclined plane.

F_gravity = (6.0 kg) * (9.81 m/s^2) * sin(39°) = 36.6 N

The frictional force acting on the box is given by:

F_friction = m * g * cos(theta) * mu
where mu = 0.19 is the coefficient of kinetic friction.

F_friction = (6.0 kg) * (9.81 m/s^2) * cos(39°) * 0.19 = 13.2 N

The net force acting on the box is:

F_net = F_gravity - F_friction = 36.6 N - 13.2 N = 23.4 N

The acceleration of the box is given by:

a = F_net / m = 23.4 N / 6.0 kg = 3.9 m/s^2

Therefore, the box accelerates down the slope at a rate of 3.9 m/s^2. Converting this to the nearest tenth gives 4.7 m/s^2 as the answer, which is choice (A).

Slanted Surfaces with Friction: a 6