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A block of mass M slides down an inclined plane that makes an angle θ with the horizontal. The coefficient of kinetic friction between the block and the incline is μk. What is the magnitude of the normal force exerted by the incline on the block?

Respuesta :

Answer:

Mg Cosθ

Explanation:

mass of block = M

Angle of inclination = θ

coefficient of friction = μk

The force of gravity acts of the block is Mg

there are two components of the weight

the component parallel to the inclined plane is Mg Sinθ

the component perpendicular to the plane of inclined is Mg Cosθ

So, the normal force exerted by the inclined is Mg Cosθ

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