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A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs of the same constant K= 250 N/m, then the block is pulled a distance A and released. What's the speed V and angular velocity w of the block as it passes through the equilibrium when the two springs are arranged in:
a) parallel
b) series

A block of mass M10 kg is on a frictionless surface as shown in the photo attached And its attached to a wall by two springs of the same constant K 250 Nm then class=

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a.

  • i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
  • ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

b.

  • i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s
  • ii. the angular velocity when the two springs are in series is 11.2 rad/s

a.

i. How to calculate the velocity of the mass when the springs are connected in parallel?

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

  • k' = equivalent spring constant in parallel = 500 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

ii. The angular velocity of mass when the springs are in parallel

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

b.

i. How to calculate the velocity of the mass when the springs are connected in series?

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

  • k" = equivalent spring constant in series = 125 N/m,
  • A = maximum displacement of spring,
  • x = equilibrium position = 0 m,
  • M = mass of block = 10 kg and
  • v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

ii. The angular velocity of the mass when the springs are in series

Since velocity of spring v = ω√(A² - x²) where

  • ω = angular velocity of spring,
  • A = maximum displacement of spring and
  • x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

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