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A block 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 greatest?---At the highest point of the oscillation.---At the new equilibrium position of the oscillation.---At the lowest point of the oscillation

Respuesta :

Answer:

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

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