13 i i) A i 5-kg object is suspended in static equilibrium from a vertical sprin
ID: 2034104 • Letter: 1
Question
13 i i) A i 5-kg object is suspended in static equilibrium from a vertical spring, and so causes the spring to stretch by 0.25 m from its unstrained length. (a) Find the spring constant of the spring. (b) The object is pulled straight down by an additional distance of 0.15 m and released from rest so it moves upward in the beginning of an oscillation. Find the speed with which the object passes through its original position (where it was hanging in static equilibrium) on the way up. As always, show a clear step-by- step solution with standard symbolic notation.Explanation / Answer
In static equilibrium Fnet = 0
Fe - Fg = 0
K*dx - m*g = 0
spring constant K = mg/dx
spring constant K = (1.5*9.8)/0.25 = 58.8 N/m <<----------ANSWER
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amplitude A = 0.15 m
at the position 0.15 , the spring mass system is at rest
it has potnetial energy U = (1/2)*k*A^2
at the static postion ( equilibrium ) the system has kinetic energy K = (1/2)*m*v^2
from energy conservation
total energy is constant
PE = KE
(1/2)*kA^2 = (1/2)*m*v^2
v = A*sqrt(k/m)
speed v = 0.15*sqrt(58.8/1.5) = 0.94 m/s <<<--------ANSWER
DONE please check the answer. any doubts post in comment box
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