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A box slides down a 28.0 degree ramp with an acceleration of 1.25 m/s2. Determine the coefficient of kinetic friction between

Sagot :

okpalawalter8

Answer:

[tex]\mu=0.39[/tex]

Explanation:

From the question we are told that:

Angle [tex]\theta=28[/tex]

Acceleration [tex]a=1.25m/s^2[/tex]

Generally the equation for Frictional force  is mathematically given by

[tex]F=\muN[/tex]

Where

[tex]N=mgcos \theta[/tex]

[tex]N=mgcos 28[/tex]

Since

Friction force is acting against move of box

Therefore

[tex]mgsin(28) - 1.25m = \mu mgcos(28)[/tex]

[tex]\mu=\frac{gsin(28) - 1.25}{gcos(28)}[/tex]

[tex]\mu=0.39[/tex]

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