## 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
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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

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2000000000 2. An object of mass 1.50 kg and density 3500 kg/m is completely submerged under oil and in a 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/m®). a) Draw a free body diagram for the suspended object. P. b) Derive the algebraic expression for the stretched distance in the spring. c) Calculate the stretched distance in the spring.

2000000000 2. An object of mass 1.50 kg and density 3500 kg/m is completely submerged under oil and in a 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/m®). a) Draw a free body diagram for the suspended object. P. b) Derive the algebraic expression for the stretched distance in the spring. c) Calculate the stretched distance in the spring.

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Fluid Pressure

The term fluid pressure is coined as, the measurement of the force per unit area of a given surface of a closed container. It is a branch of physics that helps to study the properties of fluid under various conditions of force.

Gauge Pressure

Pressure is the physical force acting per unit area on a body; the applied force is perpendicular to the surface of the object per unit area. The air around us at sea level exerts a pressure (atmospheric pressure) of about 14.7 psi but this doesn’t seem to bother anyone as the bodily fluids are constantly pushing outwards with the same force but if one swims down into the ocean a few feet below the surface one can notice the difference, there is increased pressure on the eardrum, this is due to an increase in hydrostatic pressure.

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## Problem Statement

An 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.

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