A solid metal block with a mass of 1.30 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The block is pulled to a distance of 0.200 m from its equilibrium position, held in place with a force of 19.0 N, and then released from rest. It then oscillates in simple harmonic motion. (The block oscillates along the x-axis, where x = 0 (e) What is the maximum acceleration of the block? (Enter the magnitude in m/s2.) m/s2 (f) At what position(s) (in m) on the x-axis does the maximum acceleration occur? (g) What is the total mechanical energy of the oscillating spring-block system (inJ)? h) What is the speed of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s () What is the magnitude of the acceleration of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s2 is the equilibrium position.)

e)   a = 14.62 m / s², f)  the maximum acceleration occurs at the points x = ± A , g)    Em = 1.9 J , h)   v = 1.61 m / s, i) a = -4.82 m / s²

Explanation:

A system of a spring with a mass is described by the equation

             x = A cos (wt + Ф)

Where

           w = √ k / m

e) ask to find the acceleration, for this we use the definition of acceleration

            v = dx / dt

            a = dv / dt

            v = -A w cos (wt + Ф)

            a = -A w² cos (wt + Ф)

The acceleration is maximum when the cosine is ± 1

            a = A w²

In the exercise, indicate that the amplitude is A = 0.200 m

To find the constant k, let's use Newton's equilibrium equation

          F - Fe = 0

          F = k x

          k = F / x

          k = 19 / 0.2

          k = 95 N / m

Now we can calculate the angular velocity

          w = √ k / m

         w =√ (95 / 1.3)

         w = 8.55 rad / s

Substitutes in the maximum acceleration equation

          a = 0.2 8.55²

          a = 14.62 m / s²

f) the acceleration depends on the cosine and the cosine is maximum in the express of the displacement, therefore the maximum acceleration occurs at the points x = ± A

g) mechanical energy is given by the equation

         Em = ½ k A²

         Em = ½ 95 0.2²

         Em = 1.9 J

h) what is the speed when the position is x = 1/3 A

        Em = K + U

        Em = ½ m v² + ½ k x²

        v² = (Em - ½ k x²) 2 / m

        v² = (1.9 - ½ 95 (0.2 /3)²)  2 / 1.30

        v² = 2,598

        v = 1.61 m / s

i) block acceleration when x = A / 3

For this we must find fi and the time to reach this position

Let's look fi

As the system starts from rest the speed starts is zero

         V = A w sin (0+ Ф)

          0 = sin Ф

         Ф = 0

Now we look for the time to reach the displacement x = A / 3

           A / 3 = A cos 8.5 t

           1/3 = cos 8.5t

           8.5t = cos-1 1/3

           t = 1 / 8.5 cos-1 1/3

           t = 0.1448 s

Now we substitute in the acceleration equation

          a = - A w² cos wt

          a = - 0.2 8.5² cos (8.5 0.1448)

          a = -4.82 m / s²

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s