Finding Speed Using Conservation of Energy

[Mmm_Pasta]

Homework Statement

A massless spring has unstretched length Lo = 0.45 m and spring constant k = 122.3 N/m. A block of mass m = 1.87 kg is attached to the spring, and a student stretches the spring to a length of L = 1.1 m. Then the student releases the block and it shoots upward. What is the speed of the block when it returns to the position Lo for the first time?

Use gravity = 9.81 m/s2.

Homework Equations

KE=1/2mv²
GPE=mgh
EPE=1/2kx²

The Attempt at a Solution

Using conservation of energy, I set the energies at points. EPE when the spring is stretched is equal to KE and GPE at L_o. I got 1/2kx²=mgh + 1/2mv². I found x by doing 1.1 m - 0.45 m. Since I am trying to find v I did 1/2kx² - mgh = 1/2mv². Then I plugged in the variables. 1/2(122.3 N/m)(0.65 m)² - 1.87 kg(9.81 m/s²)(0.65 m) = 1/2(1.87 kg)v². Doing the math I got 25.835875 J - 11.924055 J = (0.935 kg)v². Which is 13.91182 J/0.935 kg = v². My answer turned out to be 3.86 m/s, which turned out to be wrong.

 

[lvslugger36]

3.85 should be correct...Do you know what your source says the answer is?

 

[Mmm_Pasta]

No, it does not have the answer listed. =(

 

[PhanthomJay]

Then who said it was wrong? Try using significant figure rule: v =3.9 m/s, maybe?

 

[Mmm_Pasta]

I entered 3.9. It's probably the program thingy I am using that is fault (second time if so). This problem is from an interactive example and is just for practice anyway.