Ablock with mass m =6.7 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.28 m. while at this equilibrium position, the mass is then given an initial push downward at v = 4.6 m/s.
the block oscillates on the spring without friction.1) what is the spring constant of the spring? answer = (234.5)2) what is the oscillation frequency? answer = (0.942)3) after t = 0.47 s what is the speed of the block? 4) what is the magnitude of the maximum acceleration of the block? 5)at t = 0.47 s what is the magnitude of the net force on the block? 6)where is the potential energy of the system the the highest point of the the new equilibrium position of the the lowest point of the oscillation

1)  k = 234.5 N / m, 2)   f = 0.9416 Hz, 3)  v = 4.3 m / s , 4) a = 27.21 m / s² , 5)  182 N, 6)  maximum at the highest point

Explanation:

This is a problem of oscillatory movement  by

         x = A cos (wt + φ )

1) we can use Hooke's law to find the spring constant

        F = k x

        k = F / x

        k = mg / x

        k = 6.7 9.8 / 0.28

        k = 234.5 N / m

2) The angular velocity is

        w = √ k / m

        w = √ 234.5 / 6.7

        w = 5.916 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.916 / 2π

        f = 0.9416 Hz

3) let's look for the terms of the oscillation equation

For t = 0 v = 4.6 m / s

Speed definition

         v = dx / dt = - Aw sin (wt + φ)

We replace

        4.6 = -A 5.916 sin  (0 + φ)

By this time the body is in the equilibrium position

        0 = A cos (0 + φ)

For this relationship to be fulfilled cos (φ) = 0 ,  the cosine function is zero for φ = π/2

With this value we can find the breadth

          4.6 = -A 5.916 sin (0 + π / 2)

          A = 4.6 / 5.916

           A = 0.7776 m

Now we can write the equation of motion

         x = 0.7776 cos (5.916t + pi / 2)

Let's develop the double angle

         cos (a + 90) = cos a cos 90 - sin 90 sin a = - sin a

With this result the equation is

         x = -A sin wt

         x = - 0.7776 sin (5.916t)

Speed is

         v = -Aw cos wt

          v = - 0.7776 5.916 cos (5.916t)

          v = - 4.6 cos (5,916t)

Let's evaluate for time t = 0.47 s, remember that the angles are in radians

           v = - 4.6 cos (5,916 0.47)

          v = 4.3 m / s

4) the acceleration is

          a = dv / dt

          a = -4.6 5.916 (- sin 5.916t)

          a = 27.2136 sin (5.916t)

The acceleration is maximum when the sine function is worth 1

          a = 27.21 m / s²

5) let's use Newton's second law

         F = m a

         F = 6.7 27.21

         F = 182.3 N

6) Mechanical energy has the formula U = mgh

In general the reference system is taken at the equilibrium point, therefore the power energy is maximum at the highest point

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

no. dot product   =0 when vecs at right angles