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Chapter 5: Problem 68

Moving through a liquid, an object of mass \(m\) experiences a resistive drag force proportional to its velocity, \(F_{\text {drag }}=-b v,\) where \(b\) is a constant. (a) Find an expression for the object's speed as a function of time, when it starts from rest and falls vertically through the liquid. (b) Show that it reaches a terminal velocity \(m g / b\)

Short Answer

Expert verified
The object's speed as a function of time is given by \[v(t) = (mg/b) (1 - e^(-bt/m))\], and it reaches a terminal velocity of \[v_{\text{terminal}} = mg/b\].

Step by step solution

01

Set up the gravity and drag force

First establish the forces acting on the object. Gravity is pulling it downwards with force \(F_{\text {gravity}} = mg\). There is also the drag force, \(F_{\text {drag}} = -bv\), resisting the motion. In this expression, negative indicates that the drag force acts against the direction of motion.
02

Set up the equation of motion

The object's motion is determined by the net force acting on it. Sum up the forces and set it equal to mass times acceleration \(\a = F_{net}/m\), according to Newton's second law. The net force acting on the object is the sum of the gravity pull and drag, \(\a = (F_{\text {gravity}} + F_{\text {drag}} )/m = (mg - bv) / m\).
03

Solve the differential equation

This is a first order linear differential equation. By substituting \(\a = dv/dt\), you get \(mdv/dt = mg - bv\), which can be rearranged to \(dv/(g-bv/m) = dt/m\). This equation can then be solved by methods of integrating factors to get the velocity as a function of time \[v(t) = (mg/b) (1 - e^(-bt/m))\]
04

Derive terminal velocity

The object will achieve a terminal velocity when the speed no longer changes with time, meaning when \(dv/dt = 0\). By setting the above velocity-time expression's derivative equal to zero, and simplifying, we get the terminal velocity \(v_{\text {terminal}} = mg/b\). This shows that the drag force equals the gravitational force, so there is no net force and it no longer accelerates.

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