A 5.00 kg block is moving at 5.00 m/s along a horizontal frictionless surface to

Question

A 5.00 kg block is moving at 5.00 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall. After the block collides with the spring, the spring is compressed a maximum distance of 0.680 meters. What is the speed of the block when the spring is compressed to only half the maximum distance?
Answer is 4.33 m/s.
I started off ok at first. PE (initial) = KE (final)

Find k of the spring.

KE = 1/2 mv^2 = 1/2(5 kg)(5^2 m/s = 62.5 J

so 62.5 J = 1/2 kx^2 = 1/2 k(.680 m)^2

k = 270.33 N/M

So at half the distance, KE = 1/2 (270.33 N/M)(.34 m)^2 = 15.63 J

Find the speed of the block at half the distance:

15.63 J = 1/2 mv^2.(wrong)

This last part is where I stumbled and got it wrong. Please take it the rest of the way. Thanks!

Explanation / Answer

I started off ok at first. PE (initial) = KE (final)

Find k of the spring.

KE = 1/2 mv^2 = 1/2(5 kg)(5^2 m/s = 62.5 J

so 62.5 J = 1/2 kx^2 = 1/2 k(.680 m)^2

k = 270.33 N/m

At half the maximum distance

According to law of conservation of energy

                                    P.E + K.E = Total energy

           (1/2)K(x/2)^2 + (1/2)mv^2 = 62.5 J

                         15.625 + 2.5 v^2 = 62.5

                                       2.5 v^2 = 62.5 - 15.625

                                       2.5 v^2 = 46.87

                                            v^2 = 18.75

                                                v = 4.33 m/s    

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