A block with mass m = 7.4 kg is attached to two springs with spring constants k

Question

A block with mass m = 7.4 kg is attached to two springs with spring constants kleft = 31 N/m and kright = 52 N/m. The block is pulled a distance x = 0.26 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)? i got 21.58

2) What is the effective spring constant of the two springs? i got 83

i need help with the rest please

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 1.05 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 1.05 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?

J

Explanation / Answer

Here ,

for the springs in parallel ,

Keq = K1 + K2

Keq = 31 + 52 = 83 N/m

1) magnitude of net force = Keq * x

magnitude of net force = 0.26 * 83

magnitude of net force = 21.6 N

2)

effective spring constant is Keq = 83 N/m

3)

period of oscillation = 2pi * sqrt(m/Keq)

period of oscillation = 2pi * sqrt(7.4/83)

period of oscillation = 1.88 s

the period of oscillation is 1.88 s

4)

time taken for the block to return to equilibrium position = period /4

time taken for the block to return to equilibrium position = 0.468 s

the time taken for the block to return to equilibrium position is 0.468 s

5)

speed at the equilibrium position = A * (2pi/T)

speed at the equilibrium position = 0.26 * (2pi/1.88)

speed at the equilibrium position = 0.87 m/s

the speed at the equilibrium position is 0.87 m/s

6)

at the equilibrium position , as x = 0

net force = 0

hence , the acceleration at equilibrium position is zero

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.