The answer is supposed to be 2.4 m/s, but I\'m not sure how to arrive at that an

Question

The answer is supposed to be 2.4 m/s, but I'm not sure how to arrive at that answer. Please give a very detailed answer!

A spring with a spring constant k = 1600 N/m is at rest on the bottom of an inclined plane. A 7.0 kg block slides down the plane and makes contact with the spring at point A. After contact, the spring is compressed to point B, 0.20 m from point A, where the speed of the block is 0 m/s. What was the speed of the block just before contact with the spring? The diagram looks like this: The answer is supposed to be 2.4 m/s, but I'm not sure how to arrive at that answer. Please give a very detailed answer!

Explanation / Answer

conservation of energy :
PEspring = PEblock + KEblock

0.5kx^2 = mg(xsin55) + 0.5mv^2
0.5*1600*0.2^2 = 7*9.8*(0.2sin55) + 0.5*7v^2
32 = 11.24 + 3.5v^2
3.5v^2 = 20.76
v^2 = 5.93
v = 2.43 m/s

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