Asked by Cheech
A 2.60-kg block is attached to a horizontal spring that has a spring constant of 203 N/m. At the instant when the displacement of the spring from its unstrained length is -0.115 m, what is the acceleration a of the object? In your answer, be sure to include the proper plus or minus sign.
Answers
Answered by
Henry
Wb = 2.6kg * 9.8N/kg = 25.48 N.
203N/1m = F/-0.115
F=-0.115*203=-23.35 N.
a = F/m = -23.35/2.60 = -8.98 m/s^2.
203N/1m = F/-0.115
F=-0.115*203=-23.35 N.
a = F/m = -23.35/2.60 = -8.98 m/s^2.
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