Problem 97 - Physics for Scientists and Engineers with Modern Physics

Question

A car starts rolling down a \(1-\mathrm{in}-4\) hill \((1-\mathrm{in}-4\) means that for each 4 \(\mathrm{m}\) traveled along the road, the elevation change is 1 \(\mathrm{m} ) .\) How fast is it going when it reaches the bottom after traveling 55 \(\mathrm{m}\) ? (a) Ignore friction. (b) Assume an effective coefficient of friction equal to \(0.10 .\)

Step-by-Step Solution

Answer

Without friction, speed is 16.43 m/s. With friction, speed is 14.93 m/s.

Step 1: Understand the Problem

The car is rolling down a slope with a gradient of 1-in-4, meaning for every 4 meters along the road, the vertical drop is 1 meter. The total distance along the road is 55 meters. We need to calculate the car's speed at the bottom both with and without friction.

Step 2: Calculate the Vertical Drop

Calculate the vertical drop using the slope ratio. The vertical drop (\(h\)) can be calculated as \(h = \frac{1}{4} \times 55 = 13.75\) meters.

Step 3: Apply Energy Conservation without Friction (Part a)

Initially, potential energy is converted into kinetic energy. The potential energy at the top is \(mgh\) and the kinetic energy at the bottom is \(\frac{1}{2}mv^2\). Set them equal: \(mgh = \frac{1}{2}mv^2\). Solve for \(v\): \(v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 13.75}\).

Step 4: Calculate Speed without Friction

Substitute \(g = 9.81\, m/s^2\) and \(h = 13.75\, m\) into the equation for \(v\): \(v = \sqrt{2 \times 9.81 \times 13.75} \approx 16.43\, m/s\).

Step 5: Include Friction in Energy Conservation (Part b)

With friction, some energy is lost. The work done by friction \(W_f = f \times d = \mu mgd\), where \(\mu\) is the coefficient of friction, \(d\) is the distance along the slope. The modified equation is: \(mgh - \mu mgd = \frac{1}{2}mv^2\).

Step 6: Calculate Work Done by Friction

Calculate the work done by friction \(W_f = 0.10 \times m \times 9.81 \times 55 = 53.955m\).

Step 7: Calculate Speed with Friction

Substitute known values in the modified equation: \(mgh - \mu mgd = \frac{1}{2}mv^2\). Cancel out \(m\) and solve for \(v\): \(9.81 \times 13.75 - 53.955 = \frac{1}{2}v^2\). Solve for \(v\): \(v \approx 14.93\, m/s\).

Key Concepts

Potential Energy

Potential energy is an energy form associated with the position of an object in a gravitational field. When you are dealing with problems like a car rolling down a hill, potential energy plays a vital role. This energy is always relative to a reference point, often taken at ground level. Potential energy can be calculated using the formula: \[ PE = mgh \] where

• \( m \) is the mass of the object,
• \( g \) is the gravitational acceleration (approximately \( 9.81 \, m/s^2 \) on Earth),
• \( h \) is the height or vertical position of the object.

In the exercise given, the car starts with a certain amount of potential energy at the top of the hill because of its elevation. As the car rolls down, this potential energy is converted into kinetic energy, the energy of motion, explaining why the car speeds up.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. When discussing a car rolling down a hill, kinetic energy is what you see when the car is moving faster and faster towards the bottom. The formula for kinetic energy is: \[ KE = \frac{1}{2}mv^2 \] where

• \( m \) is the mass of the object,
• \( v \) is the velocity of the object.

When the car starts at the top of the hill, it has maximum potential energy and minimal kinetic energy. As it rolls down, the potential energy decreases as it is converted into kinetic energy, increasing the car's speed. This highlights the principle of energy conservation, which states total energy remains constant in the absence of external forces like friction.

Friction

Friction is a force that resists the motion of objects sliding or rolling over each other. In this exercise, it acts against the car's motion as it rolls downhill, converting some of the mechanical energy into thermal energy. This conversion results in the car reaching a lower speed at the bottom than it would in a frictionless environment.The effect of friction is represented by the work done, calculated as: \[ W_f = \mu mgd \] where

• \( \mu \) is the coefficient of friction, a measure of how much frictional force exists between two surfaces,
• \( m \) is the mass of the car,
• \( g \) is gravitational acceleration,
• \( d \) is the distance traveled.

Given its role in slowing the car down, friction can significantly impact the car's final velocity. In the exercise, adding friction reduces the car's speed from approximately \( 16.43 \, m/s \) to \( 14.93 \, m/s \) at the bottom of the hill. Understanding how to factor in friction is key to solving real-world physics problems accurately.