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A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction between the block and plane is 0.17, what is the magnitude of the block’s acceleration down the plane?

Respuesta :

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

[tex]F_{n}[/tex] = Normal force on the block by the ramp

[tex]f_{k}[/tex] = kinetic frictional force

Force equation perpendicular to ramp surface is given as

[tex]F_{n} = mg Cos65[/tex]

Kinetic frictional force is given as

[tex]f_{k} = \mu F_{n}[/tex]

[tex]f_{k} = \mu mg Cos65[/tex]

Force equation parallel to ramp surface is given as

[tex]mg Sin65 - f_{k} = ma[/tex]

[tex]mg Sin65 - \mu mg Cos65 = ma[/tex]

[tex]g Sin65 - \mu g Cos65 = a[/tex]

[tex](9.8) Sin65 - (0.17) (9.8) Cos65 = a[/tex]

[tex]a = 8.2[/tex] m/s²

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