A 5 kg block begins from rest and slides down a 30o inclined plane. After 4 s, the block has a velocity of 6 m/s. What is the coefficient of friction between the plane and the block?

Respuesta :

Answer:

0.577

Explanation:

Answer:

Explanation:

Parameters given:

Angle = 30°

Mass of block = 5kg

Velocity = 6 m/s

Time = 4s

Frictional force is given in terms of Normal force as:

F = uN

Where u = coefficient of frictional force

N = normal force

Frictional force is given as:

F = mgsinθ

Normal force is given as:

N = mgcosθ

Coefficient of friction is therefore given as:

mgsinθ = u * mgcosθ

=> u = sinθ / cosθ

u = tanθ

u = tan(30°)

u = 0.577

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