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 force acting on the box down the slope can be calculated as:
F_downslope = m * g * sin(θ) = 6.0 kg * 9.81 m/s^2 * sin(39°) ≈ 37.3 N

The force of kinetic friction can be calculated as:
F_friction = μ * F_norm = μ * m * g * cos(θ) = 0.19 * 6.0 kg * 9.81 m/s^2 * cos(39°) ≈ 22.9 N

The net force acting on the box down the slope is:
F_net = F_downslope - F_friction = 37.3 N - 22.9 N ≈ 14.4 N

The acceleration of the box down the slope can be calculated as:
a = F_net / m = 14.4 N / 6.0 kg ≈ 2.4 m/s^2

Therefore, the rate at which the box accelerates down the slope is:
v = a * t ≈ 2.4 m/s^2 * t

However, the initial velocity of the box is not given, so we cannot determine the rate of acceleration without additional information. Therefore, none of the answer choices are correct.

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