1. A 8.00-kg object is hung from the bottom end of a vertical spring fastened to
ID: 1373253 • Letter: 1
Question
1. A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.80 s. Find the force constant of the spring.
2. A 0.72-kg block attached to a spring with force constant 152 N/m is free to move on a frictionless, horizontal surface as in the figure below. The block is released from rest after the spring is stretched 0.13 m.
a. At that instant, find the force on the block: magnitude (unit:N)and direction
b. At that instant, find its acceleration: magnitude (unit: m/s^2) and direction
Explanation / Answer
(1)
force constant is
k = w^2 m = ( 2pi/T)^2 m = (2 pi/ 1.80 s)^2 8 kg = 97.37 N/m
(2)
force on the block is
F = kx = ( 152 N/m) ( 0.13 m) = 19.76 N ( forward direction)
from the newton second law accleration of the block is
a = F/m = 19.76 N/ 0.72 kg = 27.44 m/s^2
opposite to the direction of the force ( revese direction)
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