spring constant?

Question

The left side of the figure shows a light (`massless') spring of length 0.350 m in its relaxed position. It is compressed to 70.0 percent of its relaxed length, and a mass M= 0.210 kg is placed on top and released from rest (shown on the right).

The mass then travels vertically and it takes 1.10 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

Answer 1

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spring compression = x = 0.3(0.350) = 0.105 m

initial speed of M as it leaves spring = Vo

Vo/g = 1.10

Vo = 1.10(9.81) = 10.8 m/s

KE of mass as it leaves spring = 1/2MVo² = (0.5)(0.210)(10.8)² = 12.25 J

SPE = KE

1/2kx² = 12.25

kx² = 12.25

k = 12.25/(0.105)² = 1111 ≈ 1110 N/m ANS

Answer 2

you just forgot to do 12.25/0.5 at the and..

but tnx, it worked!