Chapter 8: Problem 30

A mass \(m\) falls under gravity (force \(m g\) ) through a liquid whose viscosity is decreasing so that the retarding force is \(-2 m v /(1+t),\) where \(v\) is the speed of \(m .\) If the mass starts from rest, find its speed, its acceleration, and how far it has fallen (in terms of \(g\) ) when \(t=1\).

Short Answer

Expert verified

Speed at \( t = 1 \): \( g(1 - e^{-1}) \), Acceleration at \( t = 1 \): \( ge^{-1} \), Distance fallen at \( t = 1 \): \( \frac{g}{2} (1 - e^{-1}) \)

Step by step solution

- Define Forces

The forces acting on the mass are the gravitational force and the retarding force. The gravitational force is given by \( mg \) and the retarding force is given by \( -\frac{2mv}{1+t} \).

- Newton's Second Law

According to Newton's second law, we set up the equation for the net force: \[ m \frac{dv}{dt} = mg - \frac{2mv}{1+t} \]

- Simplify the Equation

Divide through the equation by the mass \( m \): \[ \frac{dv}{dt} = g - \frac{2v}{1+t} \]

- Separate Variables

Rearrange the equation to separate variables: \[ \frac{dv}{g - \frac{2v}{1+t}} = dt \]

- Integrate Both Sides

Integrate both sides from \( t = 0 \) to \( t = 1 \) and from \( v = 0 \) to \( v = v \): \[ \int_0^v \frac{dv}{g - \frac{2v}{1+t}} = \int_0^1 dt \]

- Solve Integrals

After solving the integrals we get: \[ v(t) = \frac{g(1+t)}{2} (1 - e^{-\frac{2t}{1+t}}) \]

- Evaluate Speed at \( t = 1 \)

Substitute \( t = 1 \): \[ v(1) = \frac{g(1+1)}{2} (1 - e^{-\frac{2}{1+1}}) = g (1 - e^{-1}) \]

- Acceleration at \( t = 1 \)

Calculate acceleration as \( a = \frac{dv}{dt} \). From our derived speed equation: \[ a = g - \frac{2v}{1+t} \] For \( t = 1 \) and \( v(1) = g(1 - e^{-1}) \): \[ a(1) = g - \frac{2g(1 - e^{-1})}{2} = g e^{-1} \]

- Distance Fallen When \( t=1 \)

To find the distance fallen, \(x\), integrate \(v(t)\) from \(t = 0\) to \(t = 1\): \[ x = \int_0^1 v(t) dt \] Substitute \(v(t)\): \[ x = \int_0^1 \frac{g(1+t)}{2}(1 - e^{-\frac{2t}{1+t}}) dt \]Solve the integral: \[ x = \frac{g}{2} (1 - e^{1}) - 0 = \frac{g}{2} (1 - e^{-1}) \]

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Key Concepts

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

Newton's Second Law

Newton's second law is crucial in understanding how forces affect motion. The law states that the force exerted by an object is equal to the mass of the object multiplied by its acceleration, written as:
Newton's second law equation helps us analyze the forces in this exercise by providing a relationship between the retarding force and the gravitational force on the falling mass. This allows us to set up the differential equation for the net force.

By setting up the equation:

we can begin to solve for the mass's speed, acceleration, and the distance fallen over time.

Retarding Force

A retarding force is a force that opposes the motion of an object, slowing it down. In this scenario, the retarding force is dependent on both the speed of the mass and time:

This force is influenced by the viscosity of the liquid which changes with time, represented by \(-\frac{2mv}{1+t}\).

The equation for the retarding force in this exercise incorporates these variables to accurately describe how the mass's motion is being impeded. The retarding force counterbalances the gravitational force, meaning the overall motion is a result of these opposing forces.

Understanding the retarding force is essential, as it affects the velocity and acceleration of the mass over time, leading to a more complex yet comprehensive motion model.

Integration

Integration is a vital mathematical tool that allows us to solve differential equations and find functions that describe physical quantities. In this exercise, we use integration to determine the mass's velocity and distance fallen over time.

After writing down the differential equation:
we separate the variables

and then integrate both sides from the given limits:

Solving these integrals provides the functions for speed \(v\) and later for the distance \(x\).

Integration helps us understand how the variables change over time, giving us a comprehensive picture of the mass's motion through the liquid.

By integrating the speed function, we can find the distance fallen by the mass, emphasizing how integral calculus helps bridge the gap between instantaneous rates of change (like velocity) and cumulative quantities (like distance).