A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs

Question

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

Explanation / Answer

1) The spring constant K is computed with the information known about the mass at rest:

F = kx = m*g = k*0.28

k = m*g/0.28 = 6.7*9.8/0.28 = 234.5 N/m

2) The frequency of oscillation is: f = [1/(2*)] * sqrt( k/m )

f = 0.1592*sqrt[234.5 / 6.7]

f = 0.942 Hz

4) The kinetic energy at t = 0 is:

E = (1/2)*m*v2 = 0.5*6.7*4.62 = 70.886 J

At the extreme of motion, this translates entirely into additional spring potential energy. This point also represents the maximum acceleration.

Ep = (1/2)*k*(x)2 = E

x = sqrt( 2*E / k ) = sqrt( 2*70.9 / 234.5 ) = sqrt[0.6047] = 0.78 m

The additional force of the spring is:

F = k*x = 234.5*0.78 = 182.35 N

F = m*a

a = F/m = 182.35 / 6.7

a = 27.22 m/s2  

the magnitude of the maximum acceleration of the block is 27.22 m/s2

3) The equation of motion of the block is then:
x = xo + x Sin [ 2ft ]

x = 0.28 + 0.78*Sin( 2**0.94*t)

v = dx /dt = x Cos[2ft] *2f

v (0.47) = 0.78*Cos[2*3.14*0.94*0.47]*2*3.14*0.94

v (0.47) = Cos [2.7392]*4.6 = -0.92*4.6 [Note that the argument of the Cosine is in radians.]

v(0.47) = -4.2326 m/s

This means that the mass is moving upward at 4.23 m/s.

5) According to the equation of motion, the x displacement at 0.47 s is:

x(0.47) = 0.28 + 0.78*Sin( 2**0.94*0.47) = 0.28 + [0.78*0.3576] = 0.28+0.28

x(0.47) = 0.56 m

This causes a spring force of:

F = k*x(0.47) = 234.5*0.56

F = 131.0 N

i.e. the spring is pulling the mass up. This figure includes the m*g weight of the mass.

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