In a physics lab, you attach a 0.200-kg air-track glider tothe end of an ideal spring of negligible mass and start itoscillating. the elapsed time from when the glider first movesthrough the equilibrium point to the second time it moves throughthat point is 2.60 s. find the spring's force constant.

spring constant (K) = 0.292

Explanation:

mass (m) = 0.2 kg

time = 2.6 s

find the spring constant (K)

spring constant (K) = mω^{2}

the time the glider takes to move to move from the equilibrium point and back to the equilibrium point is half the period, therefore the period = 2 x time

period (T) = 2 x 2.6 = 5.2 s

ω = 2π / T = 2π / 5.2 = 1.21

spring constant (K) = mω^{2} = 0.2 x 1.21^{2}

spring constant (K) = 0.292

explanation:

could i give you a exaplo that could you?

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.