Mass on a vertical Spring: finding maximum speed

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SUMMARY

The discussion focuses on a physics problem involving a spring with a spring constant of k = 60 N/m and a mass of m = 1.5 kg. The spring stretches by 0.245 m when the mass is hung on it. When the mass is pulled down to stretch the spring to 0.49 m and released with an initial speed of 3.7 m/s, the maximum speed of the block is calculated using the conservation of mechanical energy. The correct approach involves accounting for gravitational work, leading to the correct maximum speed calculation.

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  • Understanding of Hooke's Law and spring constants
  • Knowledge of conservation of mechanical energy principles
  • Ability to perform calculations involving kinetic and potential energy
  • Familiarity with basic physics equations related to springs and forces
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for problem-solving strategies related to energy conservation in spring systems.

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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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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?
 

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