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A 67 kg kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m m . What is the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures. View Available Hint(s) k kk = nothing N/m N/m Submit Part B Complete previous part(s) Part C Complete previous part(s) Part D Complete previous part(s) Provide Feedback Next

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opudodennis

Point of correction, spring constant is 2.3×10−2 m not 2.3×10−2 m m

Answer:

28577 N/m

Explanation:

From Hooke's law, F=kx where F is force applied, k is spring constant and x is compression

F=mg=67*9.81

[tex]k=\frac {mg}{x}[/tex]

[tex]k=\frac {67*9.81}{2.3\times 10^{-2}}=28576.95652 N/m[/tex]

Approximately, 28577 N/m

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