## Problem StatementAn object of mass 150 kg and density 3600 kg/m³ is completely submerged under oil and in static equilibrium, hanging from a spring (k = 150 N/m) that is fixed from the top as shown in the figure. The density of oil is 920 kg/m³.### Tasks:a) Draw a free body diagram for the suspended object.b) Derive the algebraic expression for the stretched distance in the spring.c) Calculate the stretched distance in the spring.---### Diagram Description:The diagram shows a container partially filled with oil. Inside the container, a spring is suspended from the top. The spring is attached to a spherical object submerged in the oil. Arrows indicate forces acting on the object. The force of gravity (mg) is acting downward, and a buoyant force (\(F_b\)) is acting upward.### Given Calculation:- \(F_b\) (Buoyant force) and \(mg\) (Gravitational force) need to be considered in equilibrium.- The spring constant, \(k\), is 150 N/m.- A calculation using the formula \( k = \frac {2(0.5)(9.8)(6.945) - (6.145)}{0.002} \) is provided. The derived equilibrium constant \(k\) comes out to be approximately 0.002.Additional calculations and derivations are handwritten beside the problem.

Name: ## Problem StatementAn object of mass 150 kg and density 3600 kg/m³ i
Uploaded: 2024-03-20T06:51:10.000Z
Channel: Bartleby
Description: ## Problem StatementAn object of mass 150 kg and density 3600 kg/m³ is completely submerged under oil and in static equilibrium, hanging from a spring (k = 150 N/m) that is fixed from the top as shown in the figure. The density of oil is 920 kg/m³.#

where, Fs=spring forceFs=-kxFb=buoyant forceFb=-ρgVwhere,k=spring constantx=displacement of springρ=density of fluidg=acceleration due to gravityV=volume of an object At equilibrium condition net force acting on an object is zero. mg+Fs+Fb=0mg-kx-ρgV=0kx=mg-ρgVx=mg-ρgVk 1) Equation-1) is algebraic expression for displacement of a spring Values given in the question are as follow; ρ=920kgm3k=150Nmg=9.8ms2Mass of an objcet,mo=1.50 kgdensity of an object,ρo=3500kgm3 Volume of an object is as follows; V=moρoV=1.503500V=4.28×10-4 m3 Now, substitute known values in equation-1); x=1.509.8-4.28×10-4×920×9.8150x=0.0723 mx=7.23 cm