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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.

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Answer:

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

batolisis

The spring's force constant ( K )  is : 0.292

Given data :

mass of glider = 0.2 kg

Time = 2.6 secs

period = 2 * 2.6 = 5.2 secs

Determine the spring's force constant

spring constant can be calculated using the formula below

K = mw² ----- ( 1 )

where : m = 0.2 kg ,  w = 2π / period

therefore; w = 2π / 5.2 = 1.21

Back to equation ( 1 )

spring constant ( K ) = 0.2 * ( 1.21 )²

                                  = 0.292

Hence we can conclude that The spring's force constant ( K )  is : 0.292

Learn more about spring force constant :https://brainly.com/question/12253978

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