Problem 35 - College Physics

Question

An out-of-control truck with a mass of \(5000 \mathrm{~kg}\) is traveling at \(35.0 \mathrm{~m} / \mathrm{s}\) (about \(80 \mathrm{mi} / \mathrm{h}\) ) when it starts descending a steep \(\left(15^{\circ}\right)\) incline. The incline is icy, so the coefficient of friction is only \(0.30 .\) Use the work-energy theorem to determine how far the truck will skid (assuming it locks its brakes and skids the whole way) before it comes to rest.

Step-by-Step Solution

Answer

The truck skids approximately 198.6 meters before stopping.

Step 1: Understanding the Problem

We need to find the distance that the truck skids before coming to a stop using the work-energy theorem. The work-energy theorem states that the work done by all forces is equal to the change in kinetic energy.

Step 2: Calculate Initial Kinetic Energy

The initial kinetic energy (KE) is given by \( KE = \frac{1}{2} m v^2 \). Here, \( m = 5000 \, \text{kg} \) and \( v = 35.0 \, \text{m/s} \). Thus, \( KE = \frac{1}{2} \times 5000 \times 35.0^2 = 3062500 \, \text{J} \).

Step 3: Determine Work Done by Friction and Gravity

Work done by gravity is \( W_{gravity} = m g d \sin \theta \), and work done by friction is \( W_{friction} = -\mu m g d \cos \theta \). The negative sign indicates that friction opposes the motion. \( g = 9.8 \, \text{m/s}^2, \mu = 0.30, \theta = 15^{\circ}, \sin 15^{\circ} \approx 0.2588 \), and \( \cos 15^{\circ} \approx 0.9659 \).

Step 4: Set Up the Work-Energy Equation

According to the work-energy theorem: \[ m g d \sin \theta - \mu m g d \cos \theta = -KE \]. Substituting the known values gives: \[ 5000 \times 9.8 \times d \times 0.2588 - 0.30 \times 5000 \times 9.8 \times d \times 0.9659 = -3062500 \].

Step 5: Solve the Equation for d

Combine the terms: \[ 5000 \times 9.8 \times d (0.2588 - 0.30 \times 0.9659) = -3062500 \]. Calculate \((0.2588 - 0.30 \times 0.9659) = -0.03149 \). Thus, \[ -0.03149 \times 5000 \times 9.8 \times d = -3062500 \].Solving for \( d \): \[ d = \frac{3062500}{0.03149 \times 5000 \times 9.8} \approx 198.6 \, \text{m} \].

Key Concepts

Understanding Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When an object is moving, it has the capacity to do work because of its velocity. The formula to calculate kinetic energy, often abbreviated as KE, is given by: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the object and \( v \) is its velocity.

• The truck in the exercise weighs \(5000 \text{ kg}\).
• It initially moves at \(35.0 \text{ m/s}\).

Applying these values, you can compute the truck's initial kinetic energy, which indicates the amount of work needed to stop the truck. Understanding kinetic energy helps us see why faster-moving objects require more force or distance to bring them to a stop. This is because the kinetic energy grows with the square of the velocity, meaning as speed doubles, kinetic energy quadruples.

Exploring Coefficient of Friction

The coefficient of friction is a measure of how much two surfaces resist sliding against each other. It is denoted by \( \mu \) and can vary depending on the materials and the condition of the surfaces involved. In this exercise, the icy incline has a coefficient of friction of only \(0.30\), which is relatively low. A lower coefficient of friction indicates that less force is needed to keep the truck moving downhill.

• A high coefficient means more resistance to motion.
• A low coefficient, like on ice, means less resistance.

The coefficient of friction plays a crucial role in determining the work done by friction as the truck skids. It's essential to understand that the coefficient is dimensionless, representing a ratio, not an actual force. Thus, it helps calculate forces but isn't one itself. This concept is vital in analyzing scenarios like braking and skidding.

Significance of the Inclined Plane

An inclined plane is a flat surface tilted at an angle, used to facilitate raising or lowering a load. In physics problems, inclines allow us to study how forces such as gravity and friction interact with moving objects. In this problem, the truck is descending a \(15^{\circ}\) incline, which affects how gravity assists the truck's motion downwards. Key points include:

• Inclines reduce the effective weight pulling against gravity.
• Angles change how force components are analyzed.

This alteration impacts the truck's motion, as gravity now partly contributes to its kinetic energy. The work done by gravity depends on the incline's angle due to the \( \sin \theta \) term in the work formula. In incline problems, gravity's component along the slope dictates the direction and magnitude of motion, emphasizing how slope angle directly influences dynamics.

Calculating Work Done by Friction

The work done by friction is an essential factor in stopping the truck. Friction opposes motion and does negative work to reduce the kinetic energy of the moving truck. It is calculated using the formula: \[ W_{\text{friction}} = - \mu m g d \cos \theta \] where:

• \( \mu \) is the coefficient of friction, \(0.30\) in this case,
• \( m \) is the mass of the truck,
• \( g \) is the acceleration due to gravity, \(9.8 \text{ m/s}^2\),
• \( d \) is the distance skidded, and
• \( \cos \theta \) accounts for the incline's angle.

Friction's work does more than just slow the truck; it transforms kinetic energy into other forms, like heat. Understanding how to calculate this work is key to predicting stopping distances on slippery surfaces. Negative work indicates energy is being removed rather than added, an important concept in energy conservation problems.