- #1

Mmm_Pasta

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## 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=1/2mv

^{2}

GPE=mgh

EPE=1/2kx

^{2}

## 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 1/2kx

^{2}=mgh + 1/2mv

^{2}. I found x by doing 1.1 m - 0.45 m. Since I am trying to find v I did 1/2kx

^{2}- mgh = 1/2mv

^{2}. Then I plugged in the variables. 1/2(122.3 N/m)(0.65 m)

^{2}- 1.87 kg(9.81 m/s

^{2})(0.65 m) = 1/2(1.87 kg)v

^{2}. Doing the math I got 25.835875 J - 11.924055 J = (0.935 kg)v

^{2}. Which is 13.91182 J/0.935 kg = v

^{2}. My answer turned out to be 3.86 m/s, which turned out to be wrong.