/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 The sports car has a mass of \(2... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The sports car has a mass of \(2.3 \mathrm{Mg}\) and accelerates at \(6 \mathrm{~m} / \mathrm{s}^{2}\), starting from rest. If the drag resistance on the car due to the wind is \(F_{D}=(10 v) \mathrm{N},\) where \(v\) is the velocity in \(\mathrm{m} / \mathrm{s},\) determine the power supplied to the engine when \(t=5 \mathrm{~s}\). The engine has a running efficiency of \(\varepsilon=0.68\).

Short Answer

Expert verified
The power supplied to the engine when \(t = 5 \, s\) is approximately 622 kW.

Step by step solution

01

Determine the velocity of the car at t = 5s

The velocity \(v\) of the car at time \(t\) can be obtained using the formula \(v = u + a \cdot t\), where \(u\) is the initial velocity, \(a\) the acceleration and \(t\) the time. The car starts from rest so \(u = 0\). Substituting the given values we get: \(v = 0 + 6 \cdot 5 = 30 \, m/s\).
02

Determine the force acting on the car at t = 5s

The force on the car is the sum of the driving force corresponding to the acceleration of the car and the drag force. The driving force is \(F_{D} = m \cdot a\), where \(m = 2.3 \times 10^3 \, kg\) (since 1 Mg = 10^3 kg) and \(a = 6 \, m/s^2\). This gives \(F_{D} = 13,800 \, N\). The drag force given by \(F_{d} = 10v\), where \(v = 30 \, m/s\), gives \(F_{d} = 300 \, N\). Hence, total force \(F = F_{D} + F_{d} = 13,800 + 300 = 14,100 \, N\).
03

Calculate the power required to overcome these forces at t = 5s

The power \(P\) needed to overcome these forces is given by the formula \(P = F \cdot v\), substituting the obtained values we find \(P = 14,100 \cdot 30 = 423,000 \, W\), or 423 kW.
04

Calculate the actual power supplied to the engine at t = 5s

The power calculated in the last step is the power output, but the engine has an efficiency of \(\varepsilon = 0.68\), hence the power input to the engine \(P_{i}\) is given by the relation \(P = \varepsilon \cdot P_{i}\). Thus, rearranging and calculating for \(P_{i}\) we get \(P_{i} = P / \varepsilon = 423,000 / 0.68 \approx 622,059 \, W\), or approximately 622 kW.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Calculation
When a car accelerates, its engine must provide enough power to overcome various forces and propel the car forward. The power required is influenced by factors like the force exerted by the engine and the vehicle's velocity. In our specific case, the formula for power is given by the product of force and velocity: \( P = F \cdot v \).
Here, force is the sum of the driving force generated by the engine and the drag force due to air resistance. The velocity is the car's speed at the time for which we are calculating the power. In practice, this means the faster the car goes, the more power is needed. This relationship is linear, so doubling the velocity doubles the power requirement.
Understanding this concept is crucial in automotive engineering and helps in optimizing engines to be power-efficient while meeting performance standards.
Force and Motion
Force and motion are key principles in dynamics. Force is any interaction that changes the motion of an object. Motion is the change in the position of an object over time. In engineering mechanics, the equation \( F = m \cdot a \) relates force (\(F\)), mass (\(m\)), and acceleration (\(a\)), illustrating that applying a force to a mass causes it to accelerate.
In the provided example, a car with a mass of 2.3 Mg (which equals 2300 kg) accelerates at \(6 \text{ m/s}^2\). This acceleration results from the engine's force minus any opposing forces, such as drag. Understanding how forces contribute to motion helps in designing systems that efficiently and effectively manage vehicle dynamics.
Drag Force
Drag force is a type of friction that occurs when an object moves through a fluid, like air. It opposes the motion and increases with the square of the velocity: \( F_{d} = c \cdot v^2 \) in many real-world applications, though in our exercise it was simplified to a linear relationship \( F_{d} = 10v \).
This force depends on various factors like the shape of the object, its speed, and the properties of the fluid. Streamlining a car or using lightweight materials can help reduce drag, improving fuel efficiency and top speed.
In the exercise, when the car reaches a velocity of 30 m/s, the drag force becomes 300 N. While this might seem small compared to the engine's force, it is crucial for understanding why engines need to overcome more resistance at higher speeds.
Efficiency of Engines
Efficiency in engines refers to how much of the input energy is converted into useful work. It is typically expressed as a percentage. A high-efficiency engine means that most of the fuel energy is converted into motion, with minimal waste.
In our example, the engine efficiency is 68% or \( \varepsilon = 0.68 \). This means that only 68% of the power generated by the fuel is used to move the car, with the rest lost to factors like heat, friction, and sound.
To find out how much actual power an engine needs to provide the output power calculated, the efficiency equation \( P = \varepsilon \cdot P_i \) is used, where \( P_i \) is the input power. By understanding and improving engine efficiency, we can reduce fuel consumption, lower emissions, and create more sustainable vehicles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The collar has a mass of \(20 \mathrm{~kg}\) and is supported on the smooth rod. The attached springs are undeformed when \(d=0.5 \mathrm{~m} .\) Determine the speed of the collar after the applied force \(F=100 \mathrm{~N}\) causes it to be displaced so that \(d=0.3 \mathrm{~m}\). When \(d=0.5 \mathrm{~m}\) the collar is at rest.

The \(12-\mathrm{kg}\) block has an initial speed of \(v_{0}=4 \mathrm{~m}\) when it is midway between springs \(A\) and \(B\). After striking spring \(B\), it rebounds and slides across the horizontal plane toward spring \(A,\) etc. If the coefficient of kinetic friction between the plane and the block is \(\mu_{k}=0.4,\) determine the total distance traveled by the block before it comes to rest.

The steel ingot has a mass of \(1800 \mathrm{~kg}\). It travels along the conveyor at a speed \(v=0.5 \mathrm{~m} / \mathrm{s}\) when it collides with the "nested" spring assembly. If the stiffness of the outer spring is \(k_{A}=5 \mathrm{kN} / \mathrm{m}\) determine the required stiffness \(k_{B}\) of the inner spring so that the motion of the ingot is stopped at the moment the front, \(C,\) of the ingot is \(0.3 \mathrm{~m}\) from the wall.

For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is \(F=\left[800\left(10^{3}\right) x^{1 / 2}\right] \mathrm{N},\) where \(x\) is in \(\mathrm{m}\), determine the car's maximum penetration in the barrier. The car has a mass of \(2 \mathrm{Mg}\) and it is traveling with a speed of \(20 \mathrm{~m} / \mathrm{s}\) just before it hits the barrier.

A motor hoists a \(60-\mathrm{kg}\) crate at a constant velocity to a height of \(h=5\) in \(2 \mathrm{~s}\). If the indicated power of the motor is \(3.2 \mathrm{~kW},\) determine the motor's efficiency.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.