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A 65-kg bicyclist rides his 8.8-kg bicycle with a speed of \(14 \mathrm{~m} / \mathrm{s}\). (a) How much work must be done by the brakes to bring the bike and rider to a stop? (b) What is the magnitude of the braking force if the bicycle comes to rest in \(3.5 \mathrm{~m}\) ?

Short Answer

Expert verified
Work done by brakes: 7232.4 J; Braking force: 2066.4 N.

Step by step solution

01

Calculate the Total Mass

First, we add the mass of the bicyclist and the bicycle to find the total mass. The total mass is found using the formula: \( m_{total} = m_{bicyclist} + m_{bicycle} \).Given: - Mass of the bicyclist \( m_{bicyclist} = 65 \, \text{kg} \) - Mass of the bicycle \( m_{bicycle} = 8.8 \, \text{kg} \)Thus, \( m_{total} = 65 + 8.8 = 73.8 \, \text{kg} \).
02

Determine the Initial Kinetic Energy

The kinetic energy of the bicycle system is given by the formula: \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the speed.Using the total mass \( m_{total} = 73.8 \, \text{kg} \) and speed \( v = 14 \, \text{m/s} \), we have:\[ KE = \frac{1}{2} \times 73.8 \times (14)^2 = \frac{1}{2} \times 73.8 \times 196 = 7232.4 \, \text{Joules} \].
03

Calculate the Work Done by the Brakes

To bring the bike and rider to a stop, the brakes must perform work equal to the initial kinetic energy of the system. Therefore, the work done by the brakes is equal to the initial kinetic energy calculated in Step 2.Thus, the work done by the brakes is \( 7232.4 \, \text{J} \).
04

Apply the Work-Energy Principle to Find the Braking Force

According to the work-energy principle, the work done by the braking force is equal to the change in kinetic energy. In this case, all kinetic energy is converted to work done to stop the bicycle system.To calculate the braking force \( F \), use the formula: \( W = F \times d \), where \( W \) is the work, and \( d \) is the distance over which the force acts, which is given as \( 3.5 \, \text{m} \).Rearranging, we have:\[ F = \frac{W}{d} = \frac{7232.4}{3.5} = 2066.4 \, \text{N} \].

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Key Concepts

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

Understanding Kinetic Energy
When we talk about kinetic energy, we're describing the energy an object possesses due to its motion. It's like the energy of movement. The faster something moves or the more massive it is, the more kinetic energy it has.
For our bicyclist and bicycle system, the kinetic energy was found using the formula:
  • Formula: \( KE = \frac{1}{2} m v^2 \), where:
    • \( m \) = total mass
    • \( v \) = velocity
Here, the movement at 14 m/s with a mass of 73.8 kg resulted in a kinetic energy of 7232.4 Joules. This amount tells us how much energy needs to be removed to bring the system to a stop.
The Work-Energy Principle Explained
The work-energy principle is a powerful concept in physics. It connects the work done by forces on an object to a change in kinetic energy. In simpler words, any work done on the system will result in a change in its energy state.
In our scenario, the work done by the brakes equated to the initial kinetic energy. This means all the movement energy reduces to zero when the bike stops. It highlights an important idea: to stop a moving object, you need to 'do work' on it equal to its kinetic energy.
Demystifying Braking Force
Braking force is the force exerted to reduce the speed of an object, like a bicycle. The strength of this force determines how quickly and efficiently you can stop.
In the exercise, the braking force needed was calculated by dividing the work done (7232.4 J) by the stopping distance (3.5 m):
  • Formula: \( F = \frac{W}{d} \), where:
    • \( W \) = work done
    • \( d \) = distance
This calculation resulted in a braking force of 2066.4 N. Such a force precisely balances the amount of energy absorbed over the stopping distance to bring everything to a rest.
Exploring Mass and Motion
Mass and motion are deeply linked in physics. The mass is the amount of matter in an object, while motion describes how fast it's moving and in what direction.
In the context of this exercise:
  • The combined mass of the rider and bicycle is crucial in determining how much energy is involved when they're moving.
  • A greater mass at the same speed means more kinetic energy.
Understanding these ideas helps us see why heavier systems require more energy to move and, therefore, more energy (or work) to stop.

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Most popular questions from this chapter

A small motor runs a lift that raises a load of bricks weighing 836 N to a height of 10.7 m in 23.2 s. Assuming that the bricks are lifted with constant speed, what is the minimum power the motor must produce?

A crow drops a \(0.11-\mathrm{kg}\) clam onto a rocky beach from a height of \(9.8 \mathrm{~m}\). What is the kinetic energy of the clam when it is \(5.0 \mathrm{~m}\) above the ground? What is its speed at that point?

You pick up a 3.4-kg can of paint from the ground and lift it to a height of \(1.8 \mathrm{~m}\). (a) How much work do you do on the can of paint? (b) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time? (c) Your friend decides not to use the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?

Analyze You throw a ball straight up into the air. It reaches a maximum height and returns to your hand. At what location(s) is the kinetic energy of the ball (a) a maximum and (b) a minimum? At what location(s) is the potential energy of the ball (c) a maximum and (d) a minimum?

Predict \& Explain The work required to accelerate a car from 0 to \(50 \mathrm{~km} / \mathrm{h}\) is \(W\). (a) Is the work required to accelerate the car from \(50 \mathrm{~km} / \mathrm{h}\) to \(150 \mathrm{~km} / \mathrm{h}\) equal to \(2 W, 3 W, 8 W\), or \(9 W\) ? (b) Choose the best explanation from among the following: A. The work to accelerate the car depends on the speed squared. B. The final speed is three times the speed that was produced by the work \(W\). C. The increase in speed from \(50 \mathrm{~km} / \mathrm{h}\) to \(150 \mathrm{~km} / \mathrm{h}\) is twice the increase in speed from 0 to \(50 \mathrm{~km} / \mathrm{h}\).

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