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Chapter 14: Problem 16

A \(3.00 \mathrm{~kg}\) object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mass of \(5.00 \mathrm{~kg}\). How far and in what direction does the object move in \(0.200 \mathrm{~s}\), assuming that it moves freely and that the drag force on it from the liquid is negligible?

Short Answer

Expert verified
The object moves upward by 0.131 meters in 0.200 seconds.

Step by step solution

01

Understand Forces Involved

First, determine the forces acting on the object. The object's weight, which acts downward, is given by \( F_{gravity} = m_{object} \cdot g \), where \( m_{object} = 3.00 \ \mathrm{kg} \) and \( g = 9.81 \ \mathrm{m/s^2} \). Additionally, the buoyant force, which acts upward and equals the weight of the displaced liquid, is \( F_{buoyant} = m_{liquid} \cdot g \), where \( m_{liquid} = 5.00 \ \mathrm{kg} \).
02

Calculate Net Force

Calculate the gravitational and buoyant forces: \[ F_{gravity} = 3.00 \cdot 9.81 = 29.43 \ \mathrm{N} \]\[ F_{buoyant} = 5.00 \cdot 9.81 = 49.05 \ \mathrm{N} \]Then, find the net force acting on the object: \[ F_{net} = F_{buoyant} - F_{gravity} = 49.05 - 29.43 = 19.62 \ \mathrm{N} \]
03

Determine Direction of Net Force

Since the buoyant force is greater than the gravitational force, the net force is directed upward, indicating the object will move upward.
04

Calculate Acceleration

Use Newton's second law \( F = m \cdot a \) to solve for acceleration \( a \): \[ 19.62 = 3.00 \cdot a \]Solving for \( a \) gives:\[ a = \frac{19.62}{3.00} = 6.54 \ \mathrm{m/s^2} \]
05

Use Kinematics to Find Displacement

Use the kinematic equation \( s = ut + \frac{1}{2} at^2 \) to find the displacement, where initial velocity \( u = 0 \), acceleration \( a = 6.54 \ \mathrm{m/s^2} \), and time \( t = 0.200 \ \mathrm{s} \):\[ s = 0 + \frac{1}{2} \times 6.54 \times (0.200)^2 \]\[ s = \frac{1}{2} \times 6.54 \times 0.04 \]\[ s = 0.1308 \ \mathrm{m} \]

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Key Concepts

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

Newton's Laws of Motion
Newton's laws of motion are fundamental principles that describe how objects respond to forces. These laws help us understand and predict the motion of objects under various conditions. Let's focus on how the first and second laws apply to our object submerged in liquid.

  • **First Law (Law of Inertia):** An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net force. In our problem, the object initially at rest begins to move because of the net upward force.
  • **Second Law:** This law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Mathematically, it is expressed as: \(F = m \cdot a\). In the exercise, the net force is the difference between the buoyant force and the gravitational force, resulting in the upward acceleration of the object.
Understanding these laws is crucial because they explain why the object accelerates upwards. The buoyant force (greater than the gravitational force) creates a net force that causes the motion. Without these principles, predicting how the object moves would be challenging.
Net Force Calculation
In this exercise, calculating the net force is a critical step. The net force determines the acceleration and direction of the object's motion.

Firstly, there are two primary forces acting on the object:
  • **Gravitational Force:** This downward force is calculated using \(F_{gravity} = m_{object} \cdot g\), where \(g\) is the acceleration due to gravity (\(9.81 \ \mathrm{m/s^2}\)). For a \(3.00 \ \mathrm{kg}\) object, this force equals \(29.43 \ \mathrm{N}\).
  • **Buoyant Force:** This upward force depends on the displaced liquid's weight, computed as \(F_{buoyant} = m_{liquid} \cdot g\). With a liquid mass of \(5.00 \ \mathrm{kg}\), the force turns out to be \(49.05 \ \mathrm{N}\).

The net force is the vector difference of these two forces:\[F_{net} = F_{buoyant} - F_{gravity} = 49.05 - 29.43 = 19.62 \ \mathrm{N}\]Because the buoyant force exceeds the gravitational force, the net force is upwards. This net force causes the object to rise in the liquid over time. Calculating it accurately ensures that we apply the correct forces to Newton's second law for finding acceleration.
Acceleration and Displacement
In exploring acceleration and displacement, we use the results from the net force calculation to find out how the object moves over time.

Once we've determined that the net force is \(19.62 \ \mathrm{N}\) upwards, we apply **Newton's second law** to find the acceleration:
  • **Acceleration Calculation:** Using the formula \(F = m \cdot a\), solve for \(a\):\[19.62 = 3.00 \cdot a \implies a = \frac{19.62}{3.00} = 6.54 \ \mathrm{m/s^2}\]

With this upward acceleration, we can calculate the object's displacement in a given time period using the kinematic equation:
  • **Kinematic Equation:** The equation \(s = ut + \frac{1}{2} at^2\) helps find how far the object travels. With an initial velocity \(u = 0\), time \(t = 0.200 \ \mathrm{s}\), and acceleration \(a = 6.54 \ \mathrm{m/s^2}\), we calculate:\[s = 0 + \frac{1}{2} \times 6.54 \times (0.200)^2 = 0.1308 \ \mathrm{m}\]

This result tells us the object moves upwards by \(0.1308 \ \mathrm{m}\) in that short span. Understanding these calculations is vital, as they connect force and motion through key physical laws, making predictions on movements possible.

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Most popular questions from this chapter

An office window has dimensions \(3.4 \mathrm{~m}\) by \(2.1 \mathrm{~m}\). As a result of the passage of a storm, the outside air pressure drops to \(0.93 \mathrm{~atm}\), but inside the pressure is held at \(1.0 \mathrm{~atm}\). What net force pushes out on the window?

What would be the height of the atmosphere if the air density (a) were uniform and (b) decreased linearly to zero with height? Assume that at sea level the air pressure is \(1.0 \mathrm{~atm}\) and the air density is \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\).

Water flows through a horizontal pipe and then out into the atmosphere at a speed \(v_{1}=23.0 \mathrm{~m} / \mathrm{s}\). The diameters of the left and right sections of the pipe are \(5.00 \mathrm{~cm}\) and \(3.00 \mathrm{~cm}\). (a) What volume of water flows into the atmosphere during a \(20.0 \mathrm{~min}\) period? In the left section of the pipe, what are (b) the speed \(v_{2}\) and (c) the gauge pressure?

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is \(50.0 \mathrm{~cm}\), and the density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3}\). Find the inner diameter.

The maximum depth \(d_{\max }\) that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of \(0.050 \mathrm{~atm}\). What is the difference in \(d_{\max }\) for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of \(1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) )?
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