Question

17. Suppose you attach the object with mass m to a vertical spring originally at rest, and let it bounce up and down. You release the object from rest at the spring’s original rest length. (a) Show that the spring exerts an upward force of 2.00mg on the object at its lowest point. (b) If the spring has a force constant of 10.0N/m and a 0.25-kg-mass object is set in motion as described, find the amplitude of the oscillations. (c) Find the maximum velocity.
Please show all of your work.

Answer 1

Answer:

See below ↓

Explanation:

Step 1 : Diagram

• In attachment

Step 2

• We choose the system to be the spring, the block, and the Earth and it is isolated
• We put all the data in the figure we have created and create a zero level (initial height) of the block to be yₓ = 0 and the final position, when it stops and moves upwards again, to be yₙ = -A
• No external forces are exerted on the system and no energy comes in or out of the system
• Hence,

⇒ ΔE = 0

⇒ Eₙ - Eₓ = 0

⇒ Eₙ = Eₓ

⇒ Kₙ + Uₙ + Pₙ = Kₓ + Uₓ + Pₓ

• Final kinetic energy is 0 at the lowest point

⇒ Uₙ + Pₙ = Uₓ + Pₓ

Step 3

• Initial potential energy is 0 [zero level = initial height]

⇒ Uₙ + Pₙ = Uₓ

• And we know that spring was originally at normal length, so initial spring energy is also 0

⇒ Uₙ + Pₙ = 0

⇒ 1/2kxₙ² + mgyₙ = 0

⇒ 1/2kxₙ² = -mgyₙ

• We know xₙ = A and yₙ = -A from the diagram

⇒ 1/2kA² = -mg(-A)

⇒ 1/2kA² = mgA

⇒ [1/2kA = mg]

Step 4

• Spring force is given by : F = -kx
• Note : x = A

⇒ F = kA

⇒ k = F/A

⇒ Plug 'k' into the equation found at the end of Step 3

⇒ 1/2(F/A)(A) = mg

⇒ 2F = mg

⇒ F = 2mg (a)

Step 5

• We know the spring will stop oscillating and be at rest at the new equilibrium position of the system

⇒ F - mg = 0

⇒ F = mg

• We know that :

⇒ F = -kx

• In the case x = yₙ

⇒ kyₙ = mg

⇒ yₙ = mg/k

⇒ yₙ = 0.25 x 9.8 / -10

⇒ yₙ = -0.245 m

⇒ yₙ = A

⇒ yₙ = 0.245 m (b)

Step 6

• v(max) = Aω
• v(max) = A√k/m
• v(max) = 0.245 x √(10/0.25)
• v(max) = 1.55 m/s (c)