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Chapter 8: Problem 79

A \(1500 \mathrm{~kg}\) car begins sliding down a \(5.0^{\circ}\) inclined road with a speed of \(30 \mathrm{~km} / \mathrm{h}\). The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled \(50 \mathrm{~m}\) along the road, its speed is \(40 \mathrm{~km} / \mathrm{h}\). (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

Short Answer

Expert verified
Energy reduced by 46868.16 J; frictional force magnitude is 937.36 N.

Step by step solution

01

Convert Speeds to m/s

First, convert the speeds from km/h to m/s for easier calculations. Use the conversion formula: \ \( \text{Speed in m/s} = \text{Speed in km/h} \times \frac{1000}{3600} \). \ So, \(30 \text{ km/h} = 30 \times \frac{1000}{3600} = 8.33 \text{ m/s}\). \ Similarly, \(40 \text{ km/h} = 40 \times \frac{1000}{3600} = 11.11 \text{ m/s}\).
02

Calculate Initial and Final Kinetic Energy

The kinetic energy (KE) of the car is given by \( KE = \frac{1}{2}mv^2 \). \ Calculate initial KE: \ \( KE_i = \frac{1}{2} \times 1500 \times (8.33)^2 = 52083.75 \text{ J} \). \ Calculate final KE: \ \( KE_f = \frac{1}{2} \times 1500 \times (11.11)^2 = 92550.75 \text{ J} \).
03

Calculate the Change in Potential Energy

The change in potential energy (PE) is given by \( \Delta PE = mgd \sin \theta \). \ \( m = 1500 \), \( g = 9.8 \text{ m/s}^2 \), \( d = 50 \text{ m} \), \( \theta = 5^\circ \). \ \( \Delta PE = 1500 \times 9.8 \times 50 \times \sin(5^\circ) \approx 6401.16 \text{ J}\). Note: Here, \( \sin(5^\circ) \approx 0.08716 \).
04

Calculate the Work Done by Friction

The work done by friction is the reduction in mechanical energy, which is the sum of the change in kinetic energy and the change in potential energy. \ Change in kinetic energy: \( 92550.75 - 52083.75 = 40467 \text{ J} \). \ Therefore, mechanical energy reduction is \( -6401.16 + 40467 = -46868.16 \text{ J} \). \ (Negative sign indicates energy lost due to friction.)
05

Calculate Magnitude of Frictional Force

The work done by friction \( W = F_f \times d \), where \( F_f \) is the frictional force and \( d \) is the distance traveled. \ \( -46868.16 \text{ J} = F_f \times 50 \text{ m} \). \ Solving for \( F_f \) gives: \( F_f = \frac{-46868.16}{50} = -937.36 \text{ N} \). \ The magnitude of the frictional force is \( 937.36 \text{ N} \).
06

Summary

The mechanical energy of the car is reduced by 46868.16 J due to the net frictional force. The magnitude of the net frictional force is 937.36 N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is an essential concept in physics that describes the energy a body possesses due to its motion. It's quantified using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) represents mass and \( v \) is velocity. In our car sliding down the inclined plane, kinetic energy characterizes how much energy the car has as it moves. Initially, the car started at 30 km/h, which converts to 8.33 m/s, giving it a kinetic energy of 52083.75 J. Later, with increased speed to 40 km/h (or 11.11 m/s), its kinetic energy rose to 92550.75 J.
This increase in kinetic energy shows that work was done by external forces acting on the car.
  • As velocity increases, kinetic energy quickly escalates because velocity is squared in the equation.
  • It's this change in kinetic energy that plays a key role in understanding how other forces, like friction, affect the car.
Frictional Force
Frictional force is the resistance force that acts opposite to the direction of motion, caused by the contact between two surfaces. In the context of our car on the incline, friction is the force responsible for reducing the mechanical energy. While gravity accelerates the car downward, friction opposes this motion, resulting in energy loss.
To find the work done by friction, which is equal to the change in mechanical energy, we subtract the change in potential energy from the change in kinetic energy. After calculations, this amounts to \(-46868.16 \text{ J}\), indicating energy lost due to friction.
  • Friction not only slows down or stops motion but also transforms kinetic energy into heat.
  • In this problem, the magnitude of the net frictional force is determined by \( F_f = 937.36 \text{ N} \).
Potential Energy
Potential energy refers to the energy stored due to an object's position relative to a gravitational field. For our car, this means the energy based on its height on the incline. The formula \( \Delta PE = mgd \sin \theta \) calculates the change in potential energy as the car slides down.
  • Potential energy decreases as the car loses height, converting into kinetic energy as it moves faster.
  • Here, the potential energy reduction equals 6401.16 J, signifying how much energy was accessible for conversion into movement.
Understanding potential energy helps us see how gravitational forces work in energy preservation and transformation.
Mechanical Energy
Mechanical energy is the sum of an object's kinetic and potential energy. It provides a holistic view of the energy changes occurring within a system, such as our car example. Initially, the balance between kinetic and potential energy was different from after the car traveled down the road. From conservation principles, although some mechanical energy is transformed into thermal energy through friction, it aids in calculating energy changes.
In our exercise, the mechanical energy reduction of \( 46868.16 \text{ J} \) illustrates the friction impact. This calculation confirms that all energy transitions are consistent with the principles of energy conservation.
  • Mechanical energy is integral in predicting how systems behave over time, especially in understanding speed, force, and energy loss.
  • Recognizing energy changes helps identify forces that aren't apparent鈥攍ike friction鈥攁nd assess their quantitative influence on motion.

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Most popular questions from this chapter

A factory worker accidentally releases a \(180 \mathrm{~kg}\) crate that was being held at rest at the top of a ramp that is \(3.7 \mathrm{~m}\) long and inclined at \(39^{\circ}\) to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is \(0.28\). (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s}\) ? (c) What average power does the sprinter generate during the \(1.6 \mathrm{~s}\) interval?

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of \(180 \mathrm{~N}\). The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of \(20.0 \mathrm{~cm}\) and rotates at \(2.50 \mathrm{rev} / \mathrm{s}\). The coefficient of kinetic friction between the wheel and the tool is \(0.320 .\) At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

The surface of the continental United States has an area of about \(8 \times 10^{6} \mathrm{~km}^{2}\) and an average elevation of about \(500 \mathrm{~m}\) (above sea level). The average yearly rainfall is \(75 \mathrm{~cm}\). The fraction of this rainwater that returns to the atmosphere by evaporation is \(\frac{2}{3}\); the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water-Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)
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