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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?

User Gjaa
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1 Answer

4 votes

Answer:

8.2 m/s²

Step-by-step explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17


F_(n) = Normal force on the block by the ramp


f_(k) = kinetic frictional force

Force equation perpendicular to ramp surface is given as


F_(n) = mg Cos65

Kinetic frictional force is given as


f_(k) = \mu F_(n)


f_(k) = \mu mg Cos65

Force equation parallel to ramp surface is given as


mg Sin65 - f_(k) = ma


mg Sin65 - \mu mg Cos65 = ma


g Sin65 - \mu g Cos65 = a


(9.8) Sin65 - (0.17) (9.8) Cos65 = a


a = 8.2 m/s²

A block is sliding down an inclined plane that makes an angle of 65o with the horizontal-example-1
User ZeissS
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