Problem 73 - University Physics with Modern Physics

Question

CALC Varying Coefficient of Friction. A box is sliding with a speed of 4.50 \(\mathrm{m} / \mathrm{s}\) on a horizontal surface when, at point \(P\) it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at \(P\) and increases linearly with distance past \(P\) , reaching a value of 0.600 at 12.5 \(\mathrm{m}\) past point \(P .\) (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100\(?\)

Step-by-Step Solution

Answer

(a) The box slides a distance of approximately 7.15 m. (b) Coefficient of friction at the stopping point is about 0.386. (c) With constant friction 0.100, the box slides approximately 10.245 m.

Step 1: Understand the Problem

We have a box sliding on a surface where the coefficient of friction varies linearly. We need to apply the work-energy theorem to determine the stopping distance, find the coefficient of friction at the stopping point, and compare it with a scenario where the friction coefficient is constant.

Step 2: Define the Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. For the box, the initial kinetic energy, \( KE_i = \frac{1}{2}mv^2 \), and the work done by friction \( W = -\int F_{friction} \, dx \).

Step 3: Express Friction Force as a Function of Distance

The friction force \( F_{friction} \) can be written as \( \mu(x) \cdot m \cdot g \), where \( \mu(x) \) is the linear function of the friction coefficient. Given that \( \mu(x) = 0.100 + \frac{0.600 - 0.100}{12.5}x = 0.100 + 0.04x \).

Step 4: Calculate Work Done by Friction

Using \( W = -\int_0^d (0.100 + 0.04x)mg \, dx \), we calculate the integral to find the work done over distance \( d \). This results in \( W = -mg \left( 0.100d + 0.02d^2 \right) \).

Step 5: Apply the Work-Energy Theorem

Setting \( \frac{1}{2}mv^2 = mg(0.100d + 0.02d^2) \), solve for \( d \). For \( v = 4.50 \text{ m/s} \), simplify the equation: \[ \frac{v^2}{2g} = 0.100d + 0.02d^2 \]. Substitute \( g = 9.8 \text{ m/s}^2 \) to find \( d \).

Step 6: Solve the Quadratic Equation for Distance

The equation becomes \( 1.0306 = 0.100d + 0.02d^2 \). Rearrange to form \((0.02d^2 + 0.100d - 1.0306 = 0)\) and solve the quadratic equation for \(d\). This yields the solution for \(d\).

Step 7: Determine Friction Coefficient at Stopping Point

Using the linear function \( \mu(x) = 0.100 + 0.04x \), substitute the stopping distance \( d \) back into the equation to find the coefficient of friction at the stopping point.

Step 8: Constant Friction Scenario

For part (c), consider a constant coefficient of friction \( \mu = 0.100 \). Apply the work-energy theorem again to get \( \frac{1}{2}mv^2 = \mu mgd \) and solve for new \( d \).

Key Concepts

Varying Coefficient of Friction

In this exercise, the box is subject to a coefficient of friction that changes as it slides along the rough section. Normally, we might assume the coefficient of friction to be a constant value. However, in cases of varying coefficients, it can change with respect to the position. This scenario is modeled with linear equations where the friction coefficient starts at a specific value, like 0.100 at point \( P \), and increases gradually, reaching 0.600 after 12.5 meters.

This variation is expressed as a linear function \( \mu(x) = 0.100 + 0.04x \), where \( x \) is the distance from point \( P \). Understanding this linear increase is crucial for calculating the work done by the friction force over a specific distance. It impacts the amount of kinetic energy lost by the box and ultimately determines how far it will slide before coming to a stop.

This real-world scenario accurately reflects situations where surfaces become progressively rougher, such as tires meeting increasingly icy patches on a road, or machinery that gradually wears out surfaces during operation.

Kinetic Energy

Kinetic energy is a fundamental concept that represents the energy an object possesses due to its motion. It's calculated as \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. In our scenario, the box initially has kinetic energy because it is moving at 4.50 m/s. This energy allows the box to slide along the surface until it is completely dissipated by the work done against friction.

In applying the work-energy theorem, which relates work to changes in kinetic energy, we see how the object's initial kinetic energy gets converted into work done by the friction forces against the box. As the box slows down, its kinetic energy decreases until it reaches zero at the stopping point. The energy transfer process from kinetic energy to work done by friction is an essential aspect of dynamics and energy conservation in physics.

Grasping these processes helps to understand energy transformations in various physical applications, such as brakes in vehicles which transform kinetic energy into thermal energy to stop wheels.

Friction Force

Friction force is the resistance an object encounters when moving over a surface. It's essential to estimate this force since it significantly influences the object's motion. The friction force is given by \( F_{friction} = \mu(x) \cdot mg \), where \( \mu(x) \) is the coefficient of friction dependent on the position.

In this problem, as the box travels over a rough section, its frictional resistance increases according to the linear function of the coefficient of friction. This increase opposes the box's motion, slowing it down. Understanding how friction force works in varying conditions helps one model and predict the motion of objects moving over non-uniform surfaces.

Frictional forces are everywhere, from the soles of our shoes gripping the ground to the way a car's tires interact with different surface textures on a drive.

Quadratic Equation

Solving a quadratic equation is a mathematical skill vital when dealing with problems of motion and energy, like the one presented here. The work-energy theorem provides us with an equation involving the box's kinetic energy and the work done by friction, leading to a quadratic equation of the form \( 0.02d^2 + 0.100d - 1.0306 = 0 \), where \( d \) represents distance.

To solve it, we use the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In this specific case, \( a = 0.02 \), \( b = 0.100 \), and \( c = -1.0306 \). Solving the equation gives us the distance the box slides before it completely stops.

This process not only illuminates its mathematical resolution but underscores the relationship between the equations of motion and algebraic solutions. Solving such equations is crucial for fields like physics and engineering, where modeling real-world phenomena often involve polynomial equations.