A mass (m = 3 kg) is connected to a spring of constant k (k = 50 N/m), where the

Question

A mass (m = 3 kg) is connected to a spring of constant k (k = 50 N/m), where the mass is connected at the top to a spring on a fixed incline. The coefficient of static friction between the mass and the incline is µ(subscript s) = 0.6 and the angle that the incline makes with the horizontal is ? = 20º. The mass is initially held in place so that the spring is at its natural length.

By what maximum amount can the mass be pushed down the incline and released and still have the mass in static equilibrium? Write your final answer as x(subscript s) = ___ centimeters.

Explanation / Answer

Let us push the mass down the spring by a length Xs, such that Xs is the maximum length that will allow the mass to remain in static equilibrium on the incline

Forces acting down the incline = gravity
Forces acting up the incline = friction + force due to spring

Therefore, in the state of static equilibrium:
Force due to gravity = Friction + force due to spring

Now:
Force due to gravity = mass x acceleration due to gravity x sin ( = angle of incline = 20)
Friction = µ x acceleration due to gravity x height x cos
Force due to spring = spring constant x length of deformation

Therefore:
3 x 9.8 x sin 20 = 0.6 x 9.8 x cos 20 + 50 x Xs
=> 10.055 = 5.525 + 50Xs
=> 50Xs = 4.53
=> Xs = 0.0906 m = 9.06 cm

Thus, the maximum amount can the mass be pushed down the incline and released and still have the mass in static equilibrium Xs = 9.06 cm

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