Question

n 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

Answer 1

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:

$mg sin \theta - \mu N = ma$ (1)

where

$mg sin \theta$ 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, $\theta=39^{\circ}$ being the angle of the incline

$\mu N$ is the frictional force, with $\mu = 0.6$ 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

$N-mg cos \theta =0$

where $mg cos \theta$ is the component of the weight in the direction perpendicular to the slope. Solving for N,

$N=mg cos \theta$

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

$a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2$