Homework Answers
when x = 200 mm, length of the spring in the figure, L = sqrt(90^2 + 200^2)
= 219 mm
extension of the spring, delta_x = 219 - 100
= 119 mm
Force exerted by the spring, F = k*delta_x
= 1.2*119
= 142.8 N
angle made by this force horizontal, theta = tan^-1(90/200)
= 24.2 degrees.
initial acceleration, a = Fnetx/m
= F*cos(24.2)/m
= 142.8*cos(24.2)/0.2
= 651 m/s^2 <<<<<<<<<---------------------Answer
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