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Chapter 9: Problem 89

A \(1.0\) -kg hollow ball with a radius of \(0,10 \mathrm{~m}\) and filled with air is released from rest at the bottom of a \(2.0-\mathrm{m}\) -deep pool of water. How high above the water does the ball shoot upward? Neglect all frictional effects, and neglect changes in the ball's motion when it is only partially submerged.

Short Answer

Expert verified
The height above the water can be calculated by using the derived equations. As you can see from the steps, the principle of conservation of energy and the work-energy theorem are the key principles used to solve this problem.

Step by step solution

01

Calculate the buoyant force

The buoyant force can be calculated using the formula: \( F_b = \rho V g \) where \( \rho \) is the density of water (\( 1000 kg/m^3 \)), \( V \) is the volume of the ball (\( 4/3 \pi r^3 \)) and \( g \) is the acceleration due to gravity (\( 9.81 m/s^2 \)).
02

Calculate the gravitational force

The gravitational force can be calculated using the formula: \( F_g = m g \) where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.
03

Calculate the net force

The net force can be calculated by subtracting the gravitational force from the buoyant force. \( F_{net} = F_b - F_g \)
04

Calculate the initial kinetic energy

The kinetic energy at the bottom of the pool is zero as the ball starts from rest. Therefore, \( KE_{initial} = 0 J \)
05

Calculate the final kinetic energy

The final kinetic energy can be calculated by using the work-energy theorem which states that the work done by the forces (which in this case is the net force in the upward direction) is equal to the change in kinetic energy. \( KE_{final}= F_{net} \times d \) where \( d \) is the distance travelled by the ball under the action of the net force.
06

Calculate the maximum height above water

Using the principle of conservation of energy, the height above the water is given by the equation \( KE_{final} = m_g_h \) where \( h \) is the height. Solving the equation for \( h \) will yield the result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
The gravitational force is one of the fundamental forces and acts on objects with mass. It pulls them toward the center of the Earth. For any object near Earth's surface, this force can be calculated with the formula:
\( F_g = m \, g \)
where \( F_g \) is the gravitational force, \( m \) is the object's mass, and \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
  • Gravitational force acts uniformly in a downward direction.
  • It is the primary force that gives us weight.
Understanding gravitational force helps explain why objects fall and why they need support to be up.
Work-Energy Theorem
The work-energy theorem bridges the concepts of work and energy. It states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it is expressed as:
\[ W = \Delta KE = KE_{final} - KE_{initial} \]
where \( W \) is the work done, and \( KE \) represents kinetic energy.
  • Work done by forces affects the object's kinetic energy.
  • This can help determine how speed changes due to applied forces.
In the exercise, the net force does work to change the kinetic energy of the ball as it rises.
Conservation of Energy
The principle of conservation of energy states that in an isolated system, energy cannot be created or destroyed, only transformed. For our context, the potential energy stored by an object when submerged in water converts into kinetic energy and back to potential energy as it leaves the water.
  • This principle allows us to track energy changes without loss.
  • Helps predict future states of a system.
In this problem, the energy at the bottom of the pool is stored as gravitational potential energy; once released, it transforms as the ball rises and eventually converts back into potential energy at the peak height above water.
Kinetic Energy
Kinetic energy is the energy of motion. It is defined as:
\( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity.
  • The faster an object moves, the more kinetic energy it possesses.
  • All objects in motion have kinetic energy.
When the ball is underwater and begins its upward journey, kinetic energy increases due to the net force from buoyancy, continuing until the ball exits the water, where this kinetic energy reaches zero again at its highest point.

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Most popular questions from this chapter

A hypodermic needle is \(3.0 \mathrm{~cm}\) in length and \(0.30 \mathrm{~mm}\) in diameter. What excess pressure is required along the necedle so that the flow rate of water through it will be \(1 \mathrm{~g} / \mathrm{s} ?\) (Use \(1.0 \times 10^{-8} \mathrm{~Pa}^{-} \mathrm{s}\) as the viscosity of water. \()\)

Whole blood has a surface tension of \(0.0 .58 \mathrm{~N} / \mathrm{m}\) and a density of \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). To what height can whole blood rise in a capillary blood vessel that has a radius of \(2.0 \times 10^{-6} \mathrm{~m}\) if the contact angle is zero?

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0-\mathrm{L}\) bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

A spherical weather balloon is filled with hydrogen until its radius is \(3.00 \mathrm{~m}\). Its total mass including the instruments it carries is \(15,0 \mathrm{~kg}\). (a) Find the buoyant force acting on the balloon, assuming the density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-15} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?
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