Answer:
(B) 1.6 m/s^2
Explanation:
The equation of the forces acting on the box in the direction parallel to the slope is:
[tex]mg sin \theta - \mu N = ma[/tex] (1)
where
[tex]mg sin \theta[/tex] 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, [tex]\theta=39^{\circ}[/tex] being the angle of the incline
[tex]\mu N[/tex] is the frictional force, with [tex]\mu = 0.6[/tex] 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
[tex]N-mg cos \theta =0[/tex]
where [tex]mg cos \theta[/tex] is the component of the weight in the direction perpendicular to the slope. Solving for N,
[tex]N=mg cos \theta[/tex]
Substituting into (1), solving for a, we find the acceleration:
[tex]a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2[/tex]
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
[tex]mgsin\theta-\muN=ma[/tex]
Where the Normal force
N=mgcos\theta
Therefore
[tex]mgsin\theta-ngcos\theta \mu=0\\\\a=g(sin\theta-\mucos\theta)\\\\a=9.8ms^(-2)sin39-0.6cos37\\\\a=1.6ms^{-2}\\\\[/tex]
Therefore
the box's rate of acceleration down the slope is
a=1.6m/s^2
For more information on this visit
https://brainly.com/question/23379286
