What is the spring's length and constant in Hooke's law with buoyancy force?

In summary, the conversation discusses a problem involving a ball with a mass of 250 g and a volume of 126 cm^3 being hung vertically from a spring, which causes the spring's length to become 0.690 m. The ball is then submerged in water with a density of 1000 kg/m^3, causing the spring's length to become 0.620 m. The task is to find the buoyancy force of the water on the ball, the spring's original length, and the constant (k) of the spring. The equations used to solve the problem include F=K(Lf-Li), Fb=Fg-Wapp, and Fb=density x volume x gravity. The correct solutions for
  • #1
vinamas
43
1

Homework Statement


A ball with a mass of 250 g and a volume of 126 cm^3 is hung vertically to a spring the springs length becomes 0.690 m.Then the ball is slowly submerged in water with a density of 1000 kg /m^3 then the spring's length becomes 0.620 m.

Find Buoyancy force of the water on the ball

What was the spring's original length

What is the constant (k) of the spring

Homework Equations



F=K(Lf-Li)
Fb=Fg-Wapp
Fb=density x volume x gravity

The Attempt at a Solution


I attempted to solve it by creating a the following system of equations
0.250 x 9.81 = K(0.690-Li) and 0.250 x 9.81 - 1000 x 126^-6 x 9.81 = K(0.620-Li)
Every time I solve it I get that K=17.65 and Li= 0.551 m
The answer sheet says that K=20.6 and Li = 0.561
Please help I suspect that my mistake is in the second equation but am not too sure
 
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  • #2
vinamas said:

Homework Statement


A ball with a mass of 250 g and a volume of 126 cm^3 is hung vertically to a spring the springs length becomes 0.690 m.Then the ball is slowly submerged in water with a density of 1000 kg /m^3 then the spring's length becomes 0.620 m.

Find Buoyancy force of the water on the ball

What was the spring's original length

What is the constant (k) of the spring

Homework Equations



F=K(Lf-Li)
Fb=Fg-Wapp
Fb=density x volume x gravity

The Attempt at a Solution


I attempted to solve it by creating a the following system of equations
0.250 x 9.81 = K(0.690-Li) and 0.250 x 9.81 - 1000 x 126^-6 x 9.81 = K(0.620-Li)
Every time I solve it I get that K=17.65 and Li= 0.551 m
The answer sheet says that K=20.6 and Li = 0.561
Please help I suspect that my mistake is in the second equation but am not too sure
I obtain your equations and your answers.
 
  • #3
Thanks!
 

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed.

2. What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.

3. How is Hooke's Law related to buoyancy?

Hooke's Law can be applied to buoyancy by considering the force of gravity acting on an object in a fluid and the upward force of the fluid on the object. This can be represented by the equation Fg = ρVg = Fb, where ρ is the density of the fluid, V is the volume of the object, g is the acceleration due to gravity, and Fb is the buoyant force.

4. What is buoyancy?

Buoyancy is the upward force exerted by a fluid on an object that is partially or fully submerged in the fluid. It is caused by the difference in pressure between the top and bottom of the object, with the greater pressure at the bottom pushing the object upwards.

5. How does the weight of an object affect its buoyancy?

The weight of an object does not directly affect its buoyancy. Instead, it is the density of the object compared to the density of the fluid that determines buoyancy. If the object is less dense than the fluid, it will float; if it is more dense, it will sink. The volume of the object also plays a role in determining its buoyancy.

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