/*! 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 72 (III) A block of mass \(m\) slid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(III) A block of mass \(m\) slides along a horizontal surface lubricated with a thick oil which provides a drag force proportional to the square root of velocity: $$ F_{\mathrm{D}}=-b v^{\frac{1}{2}} $$ If \(v=v_{0}\) at \(t=0,\) determine \(v\) and \(x\) as functions of time.

Short Answer

Expert verified
Velocity: \( v = (v_0^{1/2} - \frac{b}{2m} t)^2 \); integrate for position.

Step by step solution

01

Analyze the Force Equation

The drag force equation given is \( F_{D} = -b v^{\frac{1}{2}} \). This force acts opposite to the direction of velocity.
02

Apply Newton's Second Law

Newton's second law tells us that the sum of forces equals mass times acceleration: \( F = m \frac{dv}{dt} \). Thus, we have \( m \frac{dv}{dt} = -b v^{\frac{1}{2}} \).
03

Rearrange and Integrate the Differential Equation

Rearrange the equation to solve for \( \frac{dv}{dt} \): \( \frac{dv}{dt} = -\frac{b}{m} v^{\frac{1}{2}} \). Separate variables to integrate: \( \int \frac{1}{v^{\frac{1}{2}}} dv = -\int \frac{b}{m} dt \).
04

Integrate Both Sides

The left integral becomes \( 2v^{\frac{1}{2}} \), and the right integral is \( -\frac{b}{m} t + C \). Thus, \( 2v^{\frac{1}{2}} = -\frac{b}{m} t + C \).
05

Solve for Integration Constant

At \( t = 0 \), \( v = v_0 \). Substitute these into the equation: \( 2v_0^{\frac{1}{2}} = C \). Thus, \( C = 2v_0^{\frac{1}{2}} \).
06

Solve for Velocity as a Function of Time

Substitute \( C \) back: \( 2v^{\frac{1}{2}} = -\frac{b}{m} t + 2v_0^{\frac{1}{2}} \). Solve for \( v \): \( v =\left(v_0^{\frac{1}{2}} - \frac{b}{2m} t \right)^2 \).
07

Integrating Velocity to Find Position

Use \( v = \frac{dx}{dt} \), substitute for \( v \), and integrate: \(\int dx = \int \left(v_0^{\frac{1}{2}} - \frac{b}{2m} t \right)^2 dt\).
08

Evaluate the Position Function

Solve the integral from Step 7 to find \( x(t) \), being careful to apply any necessary initial conditions for position.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law
At the heart of understanding motion and dynamics is Newton's Second Law of Motion. This fundamental principle connects the force applied to an object and the resulting acceleration. It states that the force acting on an object is equal to the mass of the object times its acceleration, represented mathematically as \( F = m \frac{dv}{dt} \). In this equation, \( F \) is the force applied, \( m \) is the mass, and \( \frac{dv}{dt} \) is the acceleration, which is the rate of change of velocity with respect to time.
Newton’s Second Law helps us solve problems involving dynamics, like the one in this exercise, where a drag force opposes the motion of a sliding block. The drag force, proportional to the square root of velocity, impacts how the velocity changes over time. By setting up the correct equation using this law, we can explore the motion of the block under the influence of the drag force.
Differential Equations
Differential equations are mathematical tools used to describe phenomena involving rates of change. They are especially powerful in physics and engineering when analyzing dynamics, such as the changing velocity of a sliding block in this exercise. Here, we encounter a differential equation derived from Newton's Second Law: \( m \frac{dv}{dt} = -b v^{\frac{1}{2}} \).
This equation provides a relationship between velocity \( v \) and time \( t \), capturing the influence of the drag force characterized by \( -b v^{\frac{1}{2}} \). Solving this differential equation involves techniques such as separation of variables, allowing us to find a function that describes how velocity changes over time.
Integration plays a key role here, turning the equation from a rate of change into a cumulative measure. By integrating both sides, we convert the differential expression into a more manageable form, ultimately helping us understand the behavior of the block’s motion.
Velocity as a Function of Time
Finding velocity as a function of time involves solving the differential equation derived from the drag force and Newton's Second Law. We start from \( \frac{dv}{dt} = - \frac{b}{m} v^{\frac{1}{2}} \). By separating the variables and integrating both sides, we arrive at a relationship that connects velocity at any time to initial conditions: \( v = \left( v_0^{\frac{1}{2}} - \frac{b}{2m} t \right)^2 \).
This expression shows how velocity decreases over time due to the drag force, starting from an initial value \( v_0 \). As time progresses, the term \( \frac{b}{2m} t \) subtracts from the initial velocity term, indicating the continuous influence of the drag force. The square in the expression ensures that the velocity remains non-negative, aligning with physical reality.
Understanding this function is crucial not only for predicting future motion but also for gaining insight into how external forces like drag can shape the dynamics of systems. The exercise highlights how mathematical modeling, starting from fundamental physics laws, translates into practical scenarios regarding motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) A 1280 -kg car pulls a 350 -kg trailer. The car exerts a horizontal force of \(3.6 \times 10^{3} \mathrm{~N}\) against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

(III) An object of mass \(m\) is constrained to move in a circle of radius \(r\). Its tangential acceleration as a function of time is given by \(a_{\tan }=b+c t^{2},\) where \(b\) and \(c\) are constants. If \(v=v_{0}\) at \(t=0,\) determine the tangential and radial components of the force, \(F_{\text {tan }}\) and \(F_{\mathrm{R}}\), acting on the object at any time \(t>0\).

(III) Two drag forces act on a bicycle and rider: \(F_{\mathrm{D} 1}\) due to rolling resistance, which is essentially velocity independent; and \(F_{\mathrm{D} 2}\) due to air resistance, which is proportional to \(v^{2}\). For a specific bike plus rider of total mass \(78 \mathrm{~kg}\), \(F_{\mathrm{D} 1} \approx 4.0 \mathrm{~N} ;\) and for a speed of \(2.2 \mathrm{~m} / \mathrm{s}, F_{\mathrm{D} 2} \approx 1.0 \mathrm{~N}\) (a) Show that the total drag force is $$ F_{\mathrm{D}}=4.0+0.21 v^{2} $$ where \(v\) is in \(\mathrm{m} / \mathrm{s}\), and \(F_{\mathrm{D}}\) is in \(\mathrm{N}\) and opposes the motion. (b) Determine at what slope angle \(\theta\) the bike and rider can coast downhill at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\) s.

(II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every \(2.5 \mathrm{~s}\). If we assume their arms are each \(0.80 \mathrm{~m}\) long and their individual masses are \(60.0 \mathrm{~kg}\), how hard are they pulling on one another?

Consider a train that rounds a curve with a radius of \(570 \mathrm{~m}\) at a speed of \(160 \mathrm{~km} / \mathrm{h}\) (approximately \(100 \mathrm{mi} / \mathrm{h}\) ). ( \(a\) ) Calculate the friction force needed on a train passenger of mass \(75 \mathrm{~kg}\) if the track is not banked and the train does not tilt. (b) Calculate the friction force on the passenger if the train tilts at an angle of \(8.0^{\circ}\) toward the center of the curve.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.