When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm.

Answer 1

m•g=k•x₁
k= m•g/x₁=0.116•9.8/0.035= 32.5 N/m
When the spring rises 5.7 cm =0.057 m and the spring is compressed to x₂=0.022 m
F= k•x₂ =ma
a =k•x₂ /m=32.5•0.022/0.116 =6.16 m/s²

Answer 2

.116 * 9.81 = k (.035)
so
k = 32.5 N/m

F = .116*9.18+32.5*.022 = 1.14 + .715 = 1.855 N

1.855 = .116 a
a = 16 m/s^2

Answer 3

another way
spring is compressed .057 from equilibrium hanging

F = k (.057) = 32.5 * .057 = 1.85 N
a = 1.85/.116 = 16 m/s^2 again