A block with mass m = 6.8 kg is attached to two springs with spring constants k

Question

A block with mass m = 6.8 kg is attached to two springs with spring constants kleft = 36 N/m and kright = 56 N/m. The block is pulled a distance x = 0.29 m to the left of its equilibrium position and released from rest.

1) What is the magnitude of the net force on the block (the moment it is released)?

2) What is the effective spring constant of the two springs?

3) What is the period of oscillation of the block?

4) How long does it take the block to return to equilibrium for the first time?

5) What is the speed of the block as it passes through the equilibrium position?

6) What is the magnitude of the acceleration of the block as it passes through equilibrium?

7) Where is the block located, relative to equilibrium, at a time 1.14 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)

8) What is the net force on the block at this time 1.14 s? (a negative force is to the left; a positive force is to the right)

9) What is the total energy stored in the system?

PLEASE SHOW STEPS

Explanation / Answer

These springs are in parallel. This is true because they necessarily displace by the same amount (one compresses, one stretches). This is not true for springs in series.

1. F = (36N/m + 56N/m) * 0.29m = 26 N

2. keq = F / x = 26N / 0.29m = 89 N/m

3. T = 2(m/k) = 2(6.8kg / 89kg/s²) = 1.73 s

4. t = T / 4 = 0.432 s

5. PE becomes KE, or ½kx² = ½mv² ½ cancels
89N/m * (0.29m)² = 6.8kg * v²
v = 1.04 m/s

6. At the equilibrium point, acceleration is zero.

7. x(t) = -Acos(2t/T) = -0.29m*cos(2*1.14/1.73) = 0.157 m

8.dx/dt = v(t) = -1.04sin(3.63t + )(here w = 2/T that is w = 6.28/1.73 = 3.63.. so, therefore w = 3.63)
dv/dt = a(t) = -1.04cos(3.63t + )
a(1.14) = -1.04cos(3.63*1.14 + )
a(1.14) = -0.566 m/s2

9.Fnet = ma = 6.8*-0.566 = -3.850N

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