A block of mass m = 3 kg is connected to a spring of constant k = 50N/m, as show

Question

A block of mass m = 3 kg is connected to a spring of constant k = 50N/m, as shown in the figure below The coefficient of static friction between the block and the incline in mu0 = 0.6 and the angel that the incline makes with the horizontal is theta = 20degree. The block is initially held in place so that the spring is at its natural length. By what maximum amount can the block be pushed down the incline and released and still have the block in equilibrium? Write your final answer as x2 = abc in centimeters By what maximum amount can the block be pulled up the incline and released and still have the block is static equilibrium? Write your final answer as x0 = abc in centimeters

Explanation / Answer

givens: s = .6, m = 3kg, k = 50 N/m, = 20

First find the force of the block on the spring and the normal force.

    Fb = mg*sin = 10.07 N

    N = mg*cos = 27.66 N

The equatuin for the friction force is:

    Ff = s*N = 16.60 N

The equation of the spring force is:

   Fs = k*x = 50*x

a) when the block is pushed down and released, the static equillibrium is when the spring force is equal to the force of the block and the static frinction.

Fs = Fb + Ff

so,

   50*x = 10.07 + 16.60

   x = .5334m

   x = 53.34cm

b) when the block is pulled up and released, the static equillibrium is when the spring force and the force of the block are equal to the friction force.

   Fs + Fb = Ff

   50*x + 10.07 = 16.60

   x = .1306m

   x = 13.06m

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