Physics help
A 1.0 kg object is suspended from a vertical spring whose spring constant is 105 N/m. The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.
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Answers
The speed of the object through its original position is 4.07 m/s. This can be found with the equation v^2 = 2gs, where v is the velocity, g is the acceleration due to gravity (9.8 m/s^2) and s is the additional distance the object was pulled (0.25 m). Therefore, v = square root of (2 * 9.8 * 0.25) = 4.07 m/s.
