Problem 103 - Physics for Scientists and Engineers with Modern Physics

Question

(III) The coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between two surfaces is not strictly independent of the velocity of the object. A possible expression for \(\mu_{\mathrm{k}}\) for wood on wood is $$\mu_{\mathrm{k}}=\frac{0.20}{\left(1+0.0020 v^{2}\right)^{2}},$$ where \(v\) is in \(\mathrm{m} / \mathrm{s} .\) A wooden block of mass 8.0 \(\mathrm{kg}\) is at rest on a wooden floor, and a constant horizontal force of 41 \(\mathrm{N}\) acts on the block. Use numerical integration to determine and graph \((a)\) the speed of the block, and \((b)\) its position, as a function of time from 0 to 5.0 \(\mathrm{s}\) (c) Determine the percent difference for the speed and position at 5.0 \(\mathrm{s}\) if \(\mu_{\mathrm{k}}\) is constant and equal to \(0.20 .\)

Step-by-Step Solution

Answer

Numerical integration reveals speed and position changes. Percent differences quantifiably compare dynamics for variable versus constant friction.

Step 1: Define the Problem

Given is a wooden block on a wooden floor with a mass of 8.0 kg and a horizontal force of 41 N. The coefficient of kinetic friction, \( \mu_{\mathrm{k}} \), varies with velocity according to the expression \( \mu_{\mathrm{k}} = \frac{0.20}{(1 + 0.0020 v^2)^2} \). We need to find the block's speed and position over time using numerical integration.

Step 2: Establish the Dynamics

The forces acting on the block are the applied force \( F = 41 \text{ N} \) and the kinetic frictional force \( f_k = \mu_k N \), where \( N \) is the normal force, which equals mg in this scenario (where \( g = 9.8\, \text{m/s}^2 \)). The net force \( F_{\text{net}} \) is \( F - f_k \). Consequently, \( F_{\text{net}} = ma \), and thus, \[ a = \frac{F - \mu_k mg}{m} \]

Step 3: Implement Numerical Integration

Using the Euler method or another numerical method (such as Runge-Kutta), calculate the velocity \( v(t) \) and position \( x(t) \) of the block from \( t = 0 \) to \( t = 5.0 \text{ s} \). Start with initial conditions \( v(0) = 0 \) and \( x(0) = 0 \). Update the velocity and position at each small time step \( \Delta t \): \[\begin{align*}v(t + \Delta t) &= v(t) + a(t) \Delta t \x(t + \Delta t) &= x(t) + v(t) \Delta t\end{align*}\]

Step 4: Consider Constant Friction

If \( \mu_k \) is constant at 0.20, the frictional force will also be constant. Recalculate the acceleration using \( a = \frac{F - \mu_k mg}{m} \) with \( \mu_k = 0.20 \). Repeat the integration steps to find new \( v(t) \) and \( x(t) \) values for a constant friction scenario.

Step 5: Calculate Percent Differences

Calculate the percent difference between the final speeds and positions obtained with variable and constant friction. The percent difference formula is given by:\[\text{Percent Difference} = \frac{|\text{Variable Value} - \text{Constant Value}|}{\text{Average of Values}} \times 100\%\] Apply this to both velocity and position at \( t = 5.0 \text{ s} \).

Step 6: Graph the Functions

Using software or graphing tools, plot \( v(t) \) and \( x(t) \) over the interval from 0 to 5 seconds for both the variable and constant friction scenarios to visually compare the results.

Key Concepts

Kinetic Friction

When two surfaces slide against each other, a force known as kinetic friction acts to oppose the motion. This force is generally characterized by the coefficient of kinetic friction, denoted as \( \mu_k \). For many materials, \( \mu_k \) is considered constant, but this is not always the case. In the given exercise, the kinetic friction between wood on wood is velocity-dependent. The formula \( \mu_k = \frac{0.20}{(1 + 0.0020 v^2)^2} \) illustrates this relationship.
\( v \) here represents the velocity of the object, and as it increases, the denominator of the formula increases, leading to a decrease in \( \mu_k \). Hence, the frictional force is not only dependent on the normal force but also varies with speed.
Understanding kinetic friction helps in predicting how objects slide over each other and plays a crucial role in mechanical and transportation engineering. Analyzing scenarios where \( \mu_k \) is dynamic can give more accurate predictions of an object's motion.

Euler Method

The Euler method is one of the simplest and most widely used techniques for numerical integration. It's particularly useful for solving ordinary differential equations (ODEs) like those seen in motion problems.
In the context of the exercise provided, you need to determine the speed \( v(t) \) and position \( x(t) \) of a block over time. Euler's method approximates these values by using a step-by-step approach. Each small time step \( \Delta t \) represents a small segment of the overall motion.
The core idea is to start from known initial conditions. For example, the block starts at rest \( v(0) = 0 \) and \( x(0) = 0 \), and then, iteratively calculate:

• Velocity update: \( v(t + \Delta t) = v(t) + a(t) \Delta t \)
• Position update: \( x(t + \Delta t) = x(t) + v(t) \Delta t \)

The Euler method is straightforward but may accumulate errors over longer intervals or with larger \( \Delta t \). Despite its simplicity, it provides a good introduction to numerical solutions for dynamical systems.

Runge-Kutta Method

The Runge-Kutta method is an advanced technique for numerical integration, often considered more accurate than the Euler method. It's especially useful when handling problems where precision is crucial.
In the exercise, you could use Runge-Kutta to calculate the block's motion with the changing kinetic friction. While Euler might suffice for a quick estimate, Runge-Kutta provides a refined approach by considering intermediate steps within each time interval. This reduces the potential error that might build up over time.
This method utilizes several evaluations of the slope (rate of change) to determine the next value of the function. Typically, the fourth-order Runge-Kutta is used for a good balance of accuracy and computational efficiency:

• The method takes into account four derivative evaluations per step: initial, midpoint, another midpoint, and endpoint derivatives.
• It calculates the weighted average of these slopes to update the solution.

The increased computational effort generally pays off in terms of accuracy, making it ideal for physics simulations where rigorous results are required.