A6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. if the coefficient of kinetic friction is 0.6, at what rate does the box accelerate down the slope? 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 0.6, at what rate does the box accelerate down the slope?
(a) 2.1 m/s2
(b) 1.6 m/s2
(c) 1.9 m/s2
(d) 1.8 m/s2

We have that for the Question  it can be said that  the box's rate of acceleration down the slope is

a=1.6m/s^2

From the question we are told

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 0.6, at what rate does the box accelerate down the slope? 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 0.6, at what rate does the box accelerate down the slope?

(A) 2.1 m/s2

(B) 1.6 m/s2

(C) 1.9 m/s2

(D) 1.8 m/s2

Generally the equation for the forces  acting on the box along the slope  is mathematically given as

Where the Normal force

N=mgcos\theta

Therefore

Therefore

the box's rate of acceleration down the slope is

a=1.6m/s^2

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(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

(1)

where

is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, being the angle of the incline

is the frictional force, with being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

where is the component of the weight in the direction perpendicular to the slope. Solving for N,

Substituting into (1), solving for a, we find the acceleration:

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newton's first law of motion