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.
- JimLv 74 years agoFavorite Answer
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
- 4 years ago
you just forgot to do 12.25/0.5 at the and..
but tnx, it worked!