How Is Maximum Speed Calculated for a Block on a Vertical Spring?

[Dauden]

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

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.

 

[tiny-tim]

Hi Dauden!

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, KE_max - KE₀ = PE₀ -PE_max …

and you haven't mentioned gravity