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a block accelerates at 3.3 m/s^2 down a plane inclined at an angle of 27°. find the kinetic friction coefficient between the block and inclined plane. acceleration due to gravity is 9.8 m/s^2

Respuesta :

MathPhys

Answer:

0.13

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down.

Normal force N pushing perpendicular to the incline.

Friction force Nμ pushing parallel up the incline.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel down the incline:

∑F = ma

mg sin θ − Nμ = ma

mg sin θ − mgμ cos θ = ma

g sin θ − gμ cos θ = a

gμ cos θ = g sin θ − a

μ = (g sin θ − a) / (g cos θ)

Given a = 3.3 m/s² and θ = 27°:

μ = (9.8 m/s² sin 27° − 3.3 m/s²) / (9.8 m/s² cos 27°)

μ = 0.13

aksnkj

The value of the kinetic coefficient of friction is 0.1320.

Given information:

The inclination of the plane is [tex]\theta=27^{\circ}[/tex].

Acceleration of the block is [tex]a=3.3\rm\;m/s^2[/tex].

Acceleration due to gravity is [tex]9.8 \rm\;m/s^2[/tex].

Let N be the normal force and f be the friction force.

Resolve the forces along vertical (perpendicular to the inclined plane),

[tex]N=mgcos\theta[/tex] where m is the mass of the block.

Now, balance the forces in the parallel plane as,

[tex]mgsin\theta-f=ma\\mgsin\theta-\mu mgcos\theta=ma\\9.81\imes sin27-\mu 9.81\times cos27=3.3\\8.740\mu=1.1536\\\mu=0.1320[/tex]

Therefore, the value of the kinetic coefficient of friction is 0.1320.

For more details about friction, refer to the link:

https://brainly.com/question/13357196

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