/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A ball rises to the surface of a... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four time the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of (1) 2 (2) 3 (3) 4 (4) 6

Short Answer

Expert verified
The frictional force is greater than the weight by a factor of 3.

Step by step solution

01

Understanding the Forces

The forces acting on the ball include its weight, the buoyant force, and the frictional force (also known as drag). Since the ball moves with constant velocity, the net force on the ball is zero.
02

Calculating the Buoyant Force

The buoyant force is equal to the weight of the liquid displaced by the ball, which can be expressed as \( F_b = \rho_{liquid} V g \), where \( \rho_{liquid} \) is the density of the liquid, \( V \) is the volume of the ball, and \( g \) is the acceleration due to gravity.
03

Writing the Net Force Equation

For the ball rising with constant velocity, the frictional force \( F_f \) plus the weight \( W \) of the ball equals the buoyant force \( F_b \). This is expressed as \( W + F_f = F_b \).
04

Expressing the Weight of the Ball

The weight of the ball can be expressed as \( W = \rho_{ball} V g \), where \( \rho_{ball} \) is the density of the ball. Since the density of the liquid is four times the density of the ball, we have \( \rho_{liquid} = 4 \rho_{ball} \).
05

Solving for the Frictional Force

Substitute \( F_b = 4 \rho_{ball} V g \) and \( W = \rho_{ball} V g \) into \( W + F_f = F_b \), giving \( \rho_{ball} V g + F_f = 4 \rho_{ball} V g \). Simplifying, we find \( F_f = 3 \rho_{ball} V g \).
06

Understanding the Relationship

The problem states that the frictional force is greater than the weight of the ball. The weight of the ball is \( \rho_{ball} V g \), and the frictional force is \( 3 \rho_{ball} V g \). Therefore, the frictional force is greater than the weight by a factor of 3.

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Key Concepts

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

Buoyant Force
When an object is immersed in a fluid, it experiences an upward force called the buoyant force. This force acts against gravity and is equal to the weight of the fluid displaced by the object. For an object submerged in a liquid, the formula for buoyant force is \( F_b = \rho_{liquid} V g \), where \( \rho_{liquid} \) represents the density of the liquid, \( V \) is the volume of the displaced fluid, which is the same as the volume of the object if fully submerged, and \( g \) is the acceleration due to gravity.
This concept was first discovered by Archimedes and is fundamental in understanding why objects float or sink in liquids. If the buoyant force is greater than an object's weight, the object will rise to the surface; if it's less, the object will sink.
Understanding buoyant force helps in figuring out the dynamics of objects in fluids, as seen in the scenario where a ball rises to the surface of a liquid.
Density of Liquids
Density is a measure of how much mass is contained in a given volume. For liquids, density is typically given in units like kilograms per cubic meter (\( \text{kg/m}^3 \)).

When considering a ball in a liquid, the density of both materials is crucial. The density of the liquid in this problem is stated to be four times that of the ball's material. This information directly impacts the net forces acting on the ball.

A higher density liquid exerts a greater buoyant force on submerged objects because it displaces more mass, causing a larger upward force. In this context, knowing the density aids in predicting the object's behavior, specifying whether it will float, sink, or remain suspended.
Constant Velocity Dynamics
Constant velocity implies that an object’s speed and direction remain unchanged over time. In physics, when an object moves with constant velocity, the net force acting upon it is zero according to Newton's first law of motion.

In the problem, the ball rises through the liquid with constant velocity, meaning all acting forces are balanced. Specifically, the upward buoyant force, the opposing frictional force from the liquid, and the weight of the ball cancel each other out.

This balance allows us to write equations that express this zero-net-force condition, essential for solving for unknowns, like the frictional force, based on given conditions and contextual relationships.
Net Force Equation
The net force equation is a representation of all the forces acting on an object. Mathematically, it's often expressed as \( \sum F = ma \), where \( \sum F \) is the sum of forces, \( m \) is mass, and \( a \) is acceleration.

Since the ball rises with constant velocity, according to the exercise, the sum of the forces (net force) is zero, as acceleration \( a \) is zero. The equation given is \( W + F_f = F_b \), where \( W \) is the weight of the ball, \( F_f \) is the frictional force, and \( F_b \) is the buoyant force.

This equation can be manipulated based on known values, like density and volume, to derive other insights. The exercise uses this equation to deduce the relationship between frictional force and the ball's weight, highlighting the dynamics of objects moving through fluids under specific conditions.

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