What is the maximum speed of a 4.00kg mass attached to a spring with k=20.0N/m?

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The discussion revolves around calculating the maximum speed of a 4.00kg mass attached to a spring with a spring constant of 20.0 N/m, pulled 0.300m from equilibrium. The correct maximum speed is determined to be approximately 0.671 m/s, derived from the relationship between spring potential energy and kinetic energy. Initially, the poster miscalculated the speed as 0.612 m/s but later resolved the issue using the formula for spring energy and kinetic energy. The calculations confirmed that the total energy in the system is converted entirely into kinetic energy at maximum speed. The poster expressed gratitude for the assistance received during the problem-solving process.
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Homework Statement


I'm sorry to bother anyone but this question is really driving me insane. It's one of my question on my notes and I've been trying to re-solve it.
It's in oscillatory unit and apparently we have to know what the speed is.

A 4.00kg mass is an horizontal frictionless surface attached to a spring with k=20.0N/m. The mass is pulled 0.300m from a equilibrium, then let go.

Homework Equations


max speed

The Attempt at a Solution

The answer is suppose to be 0.671m/s but I keep getting 0.612m/s...

..Never mind I figured it out.
 
Last edited:
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Wspring=(1/2)kx^2
so W=.5*20*.3^2=.9

since the max speed is when all the energy is kinetic you can say the Wspring=(1/2)mv^2
so .9=.5*4*v^2

.45=v^2
sqrt(.45)=.6708 or .671
 
raindrops said:
Wspring=(1/2)kx^2
so W=.5*20*.3^2=.9

since the max speed is when all the energy is kinetic you can say the Wspring=(1/2)mv^2
so .9=.5*4*v^2

.45=v^2
sqrt(.45)=.6708 or .671

Thank you for replying, I'm very happy but I already solved it.

I just remember, when you told me, that kinetic energy is in there since it's in motion. Again, thank you and I'm very sorry.
 
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