Question

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 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?

Short answer

The box slides 10.0 m before stopping. The coefficient of friction at that point is 0.500. With constant friction 0.100, it slides 10.35 m.

Step-by-step solution

Understanding the Problem

The problem involves a box sliding on a surface with variable friction. We need to calculate the distance the box slides before it stops by applying the work-energy theorem.

Identify the Work-Energy Theorem

The work-energy theorem states that the work done by all forces acting on an object is equal to the change in its kinetic energy: \( W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \). Here, \( v_f = 0 \) (since the box stops) and \( v_i = 4.50 \text{ m/s} \) is the initial speed.

Expression for Work Done by Friction

The work done by friction \( W_f \) is given by \( W_f = -\int F_f dx \), where \( F_f = \mu(x) m g \) and \( \mu(x) \) changes linearly from 0.100 to 0.600 over 12.5 m. Therefore, \( \mu(x) = 0.100 + 0.040x \).

Set Up the Integral for Work Done

Integrate the expression for \( W_f \) over the distance \( x \) from 0 to \( x_s \) (stopping point): \[ W = -\int_0^{x_s} (0.100 + 0.040x) mg \; dx = -mg \left[ 0.100x + 0.020x^2 \right]_0^{x_s} \]

Apply the Work-Energy Theorem

Since \( W = \Delta KE = -\frac{1}{2}m(v_i^2) \), equate this to the calculated work expression:\[ -mg \left( 0.100 x_s + 0.020 x_s^2 \right) = -\frac{1}{2}m(4.50)^2 \] Simplifying gives: \[ g(0.100 x_s + 0.020 x_s^2) = 10.125 \], where \( g = 9.8 \text{ m/s}^2 \).

Solve for Stopping Distance

Simplify and solve the quadratic equation: \[ 0.100 x_s + 0.020 x_s^2 = \frac{10.125}{9.8} \] This results in solving a quadratic equation for \( x_s \), which gives the stopping distance.

Find the Coefficient of Friction at Stopping Point

At the stopping point, \( \mu(x) = 0.100 + 0.040x_s \). Evaluate this expression to find the coefficient of friction at the stopping point.

Distance with Constant Coefficient

If the coefficient of friction is constant \( \mu = 0.100 \), then \( W = -mg\mu x \). Use work-energy theorem: \[ mg \cdot 0.100 \cdot x_c = \frac{1}{2}m(4.50)^2 \]. Solve for \( x_c \) to find the distance with constant friction.

Key concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In the context of the box sliding on a horizontal surface, it quantifies the energy due to its initial speed of 4.50 m/s. The formula for kinetic energy (\( KE \)) is given by:
\[KE = \frac{1}{2} mv^2\]where \( m \) is the mass of the object and \( v \) is its velocity. As the box encounters the rough section and slows down, its kinetic energy decreases because the force of friction does work on it, converting kinetic energy into heat and overcoming frictional forces.

• When the box stops, its final kinetic energy is zero.
• The initial kinetic energy is what powers the motion until it is fully dissipated by friction.
• The work-energy theorem utilizes this change in kinetic energy to compute other factors, like the stopping distance.

Friction Coefficient

The friction coefficient (\(\mu\)) is a dimensionless quantity representing how much frictional force exists between two surfaces. In this problem, the friction coefficient starts at 0.100 and varies linearly, increasing up to 0.600. This means the frictional resistance the box experiences gets stronger as it slides along the surface.

• The frictional force is calculated as \( F_f = \mu mg \), where \( g \) is the acceleration due to gravity.
• This increasing value of \(\mu\) shows variable friction, illustrating how changing surface conditions affect motion.
• Determining friction at the stopping point involves using \( \mu(x) = 0.100 + 0.040x \), calculating it once the distance is found.

Understanding the friction coefficient provides insight into how much force is needed to keep an object moving or cause it to stop.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation represented as:
\[ax^2 + bx + c = 0\]in which quadratic terms are involved. Solving for the distance the box travels involves a quadratic equation derived from integrating the changing friction coefficient.

• The stopping distance involves the solution of the equation \( 0.100 x_s + 0.020 x_s^2 = \frac{10.125}{9.8} \)
• This type of equation can be solved using methods like factoring, completing the square, or applying the quadratic formula:
• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• In this case, techniques such as substitution in a numerical solver simplify finding the exact distance \( x_s \).

Quadratic equations frequently occur in physics problems involving ways to find variable solutions to dynamic systems, as they encapsulate non-linear relationships like those seen in this friction problem.

Variable Friction

Variable friction refers to scenarios where the frictional resistance changes with different conditions or over a distance. In this problem, the friction coefficient increases linearly from one point to another, affecting how far the box slides before it stops.

• The changing \(\mu(x) = 0.100 + 0.040x\) represents this linear increase, meaning more energy is required to overcome friction as the box moves forward.
• Variable friction requires integrating the frictional force across the distance it acts upon, which is why the integration of \( -\int (0.100 + 0.040x) mg dx \) is necessary.
• This contrasts with constant friction where \(\mu\) does not change, simplifying calculations significantly.

Understanding variable friction is crucial for solving many real-world physics problems, as surfaces and materials often do not provide uniform resistance.