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Mass on Vertical Spring

  • Thread starter Dauden
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  • #1
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Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/spring/homework/Ch-08-GPE-ME/mass_vertical_spring/7.gif [Broken]

A spring with spring constant k = 40 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 1 kg is hung gently on the end of the spring.


Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 4 m/s.

What is the maximum speed of the block?

Homework Equations


KE = (1/2)mv^2
PE = (1/2)kx^2

The Attempt at a Solution



Well, the answer for part a) they want me to use is .245 m. The way I set this up was:

(1/2)mv^2 = (1/2)mvi^2 + (1/2)kx^2

(1/2)(1)v^2 = (1/2)(1)(4^2) + (1/2)(40)(.49^2)

I did this because the kinetic energy at the maximum speed point should be equal to the potential plus the kinetic energy given by the push. When I solve for v on the left side I get 5.06 m/s but that is wrong.
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Dauden! :smile:
Well, the answer for part a) they want me to use is .245 m. The way I set this up was:

(1/2)mv^2 = (1/2)mvi^2 + (1/2)kx^2

(1/2)(1)v^2 = (1/2)(1)(4^2) + (1/2)(40)(.49^2)

I did this because the kinetic energy at the maximum speed point should be equal to the potential plus the kinetic energy given by the push. When I solve for v on the left side I get 5.06 m/s but that is wrong.
No, KEmax - KE0 = PE0 -PEmax

and you haven't mentioned gravity :wink:
 

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