Physics, 07.10.2019 18:10 thechocolatblanc

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

Answers: 3

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Answer from: catelinboog04

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

(D) 1.8 m/s2

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

$mgsin\theta-\muN=ma$

Where the Normal force

N=mgcos\theta

Therefore

$mgsin\theta-ngcos\theta \mu=0\\\\a=g(sin\theta-\mucos\theta)\\\\a=9.8ms^(-2)sin39-0.6cos37\\\\a=1.6ms^{-2}\\\\$

Therefore

the box's rate of acceleration down the slope is

a=1.6m/s^2

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Answer from: torresnoemi899

(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:

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

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