When a 0.38 kg block is suspended on a vertical spring, it causes it to stretch 3.5 cm. If the block is now pulled 7.5 cm below its equilibrium position and released, what is the speed of the block when it is 4.2 cm below the equilibrium position?

  w = 16.73 rad / s,  v = - 1.04 m / s

Explanation:

This is an exercise in oscillatory motion of a system of a spring and a mass, the angular velocity is

         w =

To calculate the spring constant we use the data that Δx = 3.5 cm = 0.035 m with a mass of m = 0.38 kg, let's write Hooke's law

          F = -kx

in this case the applied force is the weight of the body

          F = W

we substitute

          mg =  k x

          k = m g / x

         k = 0.38 9.8 / 0.035

         k = 106.4 N / m

let's find the angular velocity

         w = Ra 106.4 / 0.38

         w = 16.73 rad / s

the movement of this system is described by the expression

          ax = A cos (wt + Ф)

the quantity A is called the amplitude of the movement in this case A = 7.5 cm = 0.075 m

to encode the constant we use that the system is released from its position of maximum displacement (x = A)

we substitute

          A = A cos (0 + Ф)

         cos Ф = 1

         Ф = cos-1 1

         Ф = 0º

by substituting in the equation of motion

            x = 0.075  cos (16.73 t)

Let's find the time it takes to reach the desired position x = 4.2 cm = 0.042 m

           cos 16.73 t = 0.042 / 0.075 = 0.56

            16.73 t = cos -1 0.56

remember that angles must be expressed in radians

            t = 0.976 / 16.73

            t = 0.0583 s

body velocity is

           v =

            v = -0.075 16.73 sin 16.73t

let's calculate for the time found

           v = - 1.25475 sin (16.73  0.0583)

           v = - 1.04 m / s

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explanation:

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