Mass on a vertical Spring: finding maximum speed

In summary, the conversation is about a spring with a spring constant of 60 N/m and an unstretched length of 2.2 m attached to the ceiling. A 1.5 kg block is hung on the end of the spring, causing it to stretch by 0.245 m. In the second part of the conversation, the block is pulled down until the spring is stretched by 0.49 m, and then pushed upward with an initial speed of 3.7 m/sec. The question is asking for the maximum speed of the block. The solution involves taking into account work due to gravity and using the formula .5 mv0^2 + .5kx^2 = .5 mvf^2,
  • #1
ilyketurdles
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Homework Statement



A spring with spring constant k = 60 N/m and unstretched length of L0 = 2.2 is attached to the ceiling. A block of mass m = 1.5 kg is hung gently on the end of the spring.
1)

How far does the spring stretch?

Answer is .245 m

2)

Now the block is pulled down until the total amount the spring is stretched is twice the amount found in the above question. The block is then pushed upward with an initial speed vi = 3.7 m/sec. What is the maximum speed of the block?

Homework Equations



F spring = -kx
Wnc = change in ME[/B]

The Attempt at a Solution



I found the first part just by using force, with the force required by the spring to be equal to mg, then found x from there.

For part 2, however, I'm having problems.

I took x to be .245 m (from the first part) x 2 = .49 m

Then, assuming the forces are all conservative, and that the velocity will be maximum at the spring's relaxed position, I used the formula:

.5 mv02 + .5kx2 = .5 mvf2

Solving for Vf, I get 4.82, which is incorrect. I've also tried doubling the entire spring, in case that's what the question was asking, and was still incorrect.
 
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  • #2
So I figured it out, leaving my method of solution for anyone else who may come across this problem.

I neglected work due to gravity. So what you can do is take the stretched position in problem 1 as the equilibrium position, then use the formula I used before

.5 mv02 + .5kx2 = .5 mvf2

except x is now the amount stretched for the already stretched position, in this case, .49 m. Then solve for Vf.Is there any way to close a thread?
 

Related to Mass on a vertical Spring: finding maximum speed

1. What is the equation for finding maximum speed in a mass on a vertical spring system?

The equation for finding maximum speed in a mass on a vertical spring system is given by: v = √(k/m) * A, where k is the spring constant, m is the mass, and A is the amplitude of the oscillations.

2. How do I determine the spring constant of a vertical spring system?

The spring constant can be determined by dividing the force applied to the spring by the displacement it causes. This is known as Hooke's Law, where F = -kx, with F representing the force, k representing the spring constant, and x representing the displacement.

3. Can the maximum speed of a mass on a vertical spring system be greater than the speed at equilibrium?

Yes, the maximum speed of a mass on a vertical spring system can be greater than the speed at equilibrium. This occurs when the amplitude of the oscillations is greater than the equilibrium position, causing the mass to travel further and faster.

4. How does the mass affect the maximum speed in a vertical spring system?

The mass has an inverse relationship with the maximum speed in a vertical spring system. This means that as the mass increases, the maximum speed decreases. This can be seen in the equation v = √(k/m) * A, where m is in the denominator.

5. What factors can affect the maximum speed in a mass on a vertical spring system?

The maximum speed in a mass on a vertical spring system can be affected by the spring constant, mass, and amplitude of the oscillations. Additionally, factors such as air resistance and friction can also play a role in altering the maximum speed.

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