Physics help, I cant figure out, Ideal springs and frictionless surfaces?

Question

Alright here is my problem, 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 0.68m. What is the speed of the block when the spring is compressed to only one-half of the maximum distance?

Answer 1

First, assume the collision is elastic, so that the KE of the block is transferred 100% to the spring.
Then use
.5*m*v^2=.5*k*x^2
where m=5, v=5, x=0.68
to compute the k.
once you have the K
then use
.5*m*(5^2-v^2)=.5*k*(0.68/2)^2
and solve for v

k=270.33
v=4.33 m/s at 1/2 the compressed distance

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Answer 2

TE(I) = 1/2 mV^2; where m = 5 kg and V = 5 mps upon impact I.

TE(C/2) = 1/2 mv^2 + k delX/2; where v = the velocity you are looking for, del X is the maximum compression when v = 0, and k is the compression coefficient you need to find to work this part.

TE(C) = k delX = 1/2 mV^2 = TE(I), which says the total energy at full compression C is equal to the total energy upon impact I. Thus k = mV^2/2 delX; where everything on the RHS is given.

Now TE(C/2) = 1/2 mv^2 + k delX/2 = 1/2 mV^2 = TE(I); so that 1/2 mv^2 = 1/2 mV^2 - k delX/2 and v^2 = 2[1/2 mV^2 - k delX/2]/m and everything on the RHS is given or derived. You can do the math.

The physics is this. All the total kinetic energy at impact is converted to partly potential energy of the half compression and partly remaining kinetic energy. Before you can find the potential energy at half compression, you need k and that you can get by compressing the spring the full delX and solving for k knowing that all the spring's potential is equal to the impact kinetic energy.

Answer 3

This question is a matter of energy being converted from kinetic to potential. First you find the kinetic energy using KE = 1/2*m*v^2. KE = 12.5J. Then solve for k, PE = 1/2*k*x^2. k = 54.06 N/m. Then solve for PE again using the new distance, 0.34m. New PE = 3.125N. This PE is equal to the new KE that the block has at the new velocity. Using KE = 3.125N, solve for v. v = 1.12m/s