Question

A scientist has suspended a spring from a door frame. When an object is attached to the spring's bottom end, the spring extends downward by an additional distance, d, measuring 4.30 cm. (That downward distance is relative to the unstretched position that the spring had previously held whern in equilibrium, as shown by the horizontal dashed line in the figure below Calculate the mass of the hanging object (in kg) if the spring constant k has the value 59.5 N/m Enter a number Once the spring-plus-object system reaches equilibrium, what two forces act upon the object? Are they equal in magnitude? Which of these two forces can be computed directly from "givens" in the problem statement? Which of these two forces depends on the "unknown" quantity, the object's mass? Double-check your arithmetic and be sure that you have converted all the "givens" into SI units. kg

Answer 1

$According\,\,to\,\,Hooke's\,\,law\,\,\,\,F=kx \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....(1) \\\\ Where, F\,\,is\,\,force\,\,applied.\\ k\,\,is\,\,spring\,\,constan.\\ x\,\,is\,\,displacement.\\\\ Here, F=mg\\ m \,\,is \,\,mass\,\,of\,\,the\,\,hanging\,\,object.\\ g=9.81\,\,m/s^2\,\,is\,\,acceleration\,\,due\,\,to\,\,gravity.\\ k=59.5\,\,N/m\\ x=d=4.30\,\,cm=0.043\,\m$

$Substituting\,\,in\,\,eq.\,\,(1)\,\,\\ m\times9.81\,\,m/s^2=59.5\,\,N/m\times0.043\,\,m\\\\ \Rightarrow m=\frac{59.5\times0.043}{9.81}\,\,kg\\\\\ \Rightarrow m=0.2608\,\,kg$

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