Problem 72 - Physics for Scientists and Engineers with Modern Physics

Question

(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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( m \frac{dv}{dt} = -b v^{\frac{1}{2}} \).

Step 3: 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: \( \int \frac{1}{v^{\frac{1}{2}}} dv = -\int \frac{b}{m} dt \).

Step 4: Integrate Both Sides

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

Step 5: Solve for Integration Constant

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

Step 6: 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 \( v \): \( v =\left(v_0^{\frac{1}{2}} - \frac{b}{2m} t \right)^2 \).

Step 7: Integrating Velocity to Find Position

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

Step 8: Evaluate the Position Function

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

Key Concepts

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.