Solving for Velocity of Spring-Supported Mass

[CaptainSFS]

Homework Statement

A spring is stretched a distance of Dx = 40 cm beyond its relaxed length. Attached to the end of the spring is an block of mass m = 8 kg, which rests on a horizontal frictionless surface. A force of magnitude 20 N is required to hold the block at this position. The force is then removed.

Homework Equations

W=Fd

KE=(.5)(m)(v)^2

KE(final) - KE(initial) = W

The Attempt at a Solution

I first found work, which I calculated (20N)(.4m) = 8 J

Then I plugged it into: 8 J = (.5)(8kg)(0m/s) - (.5)(8kg)(v)^2

8 = 0 - 4(v)^2

I end up with -(sqrt(2)) or +(sqrt(2)), but neither of those are correct. Any ideas with what I am doing wrong?

 

[Galileo's Ghost]

Keep in mind that the spring exerts a force of 20 N only when the spring is stretched to 40 cm. Think about how much force the spring would exert if the block were at the equilibrium position (its relaxed position). The force is not constant during the motion from the 40 cm mark to the relaxed position.

 

[CaptainSFS]

oh, how would i calculate the velocity then, if it's not 20 N when it returns? or if i use the same formula (KE(final) - KE(initial) = W), how would I calculate that Work?

 

[Galileo's Ghost]

Consider the average force on the block when determining the Work done on it from the spring.

 

[CaptainSFS]

oh alright thanks. I see now, that was easier than I was thinking. thanks!