/*! 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 17 A 2000 -kg car mowes down a leve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 17

A 2000 -kg car mowes down a level highway under the actions of two forces: a \(1000-\mathrm{N}\) forward force exerted on the drive wheels by the road and a 950-N resistive force. Use the work-energy theorem to find the speed of the car after it has moved a distance of \(20 \mathrm{~m}\), assuming that it starts from rest.

Short Answer

Expert verified
The speed of the car after it has moved a distance of 20 m, starting from rest, is approximately \( 1 \, \text{m/s} \).

Step by step solution

01

Calculate Net Force

The car is acted upon by two forces - a forward force and a resistive force. Calculate the net force on the car as the difference between the forward force and the resistive force. Net Force \(F_{\text{net}} = F_{\text{forward}} - F_{\text{resistive}} = 1000 \, \text{N} - 950 \, \text{N} = 50 \, \text{N}\)
02

Calculate Work Done

Work done on the car is calculated as the product of the force and the distance over which it is applied. Use the formula for work done \(W = F_{\text{net}} \cdot d\), where \(d\) is the distance travelled. Here, \(d = 20 \, \text{m}\). So, \(W = 50 \, \text{N} \cdot 20 \, \text{m} = 1000 \, \text{Joules}\)
03

Apply Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. The kinetic energy change \( \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \). Given that the car starts from rest, the initial kinetic energy is 0. So, \( \Delta KE = KE_{\text{final}} = W = 1000 \, \text{Joules}\)
04

Calculate the final Speed

You can find the final speed of the car by setting the calculated work equal to the kinetic energy. The kinetic energy is given by \( \frac{1}{2} m v^2 \), where m is the mass and v is the speed. Therefore, \( \frac{1}{2} m v^2 = W \). Solve for \( v \) to get the final speed of the car: \( v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \cdot 1000 \, \text{Joules}}{2000 \, \text{kg}}}\)

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 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. Imagine rolling a ball down a hill; as it speeds down, it gains kinetic energy. In general, any moving object has kinetic energy. This energy depends on two main factors: the mass of the object and its velocity鈥攐r speed鈥攁t which it moves.To calculate kinetic energy, you use the formula:
  • \( KE = \frac{1}{2} m v^2 \)
  • Here, \( m \) represents the mass of the object, and \( v \) represents the velocity.
If an object is at rest, its kinetic energy is zero. As it starts moving, this energy increases rapidly, since velocity is squared in the formula. Thus, small increases in speed lead to large increases in kinetic energy. Understanding this is crucial for solving problems using the work-energy theorem.
Work Done
Work done refers to the energy transferred when a force moves an object over a distance. When you push a heavy box across the floor, you're doing work since you're applying force to move it. This concept is fundamental in understanding how forces affect motion and energy changes in objects.The formula to calculate work done is simple:
  • \( W = F \cdot d \cdot \cos(\theta) \)
  • \( F \) is the applied force, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of motion.
If the force is in the same direction as the motion, then the angle \( \theta \) is 0 degrees, making \( \cos(\theta) = 1 \). Hence, the simplified equation used in our exercise: \( W = F \cdot d \). This concept explains why a force has to be involved to change an object's kinetic energy.
Net Force
Net force is the overall force acting on an object when all the individual forces are combined. Imagine a game of tug of war, where two teams pull with different forces. The net force determines which direction the rope moves. By understanding this concept, you can predict the motion of objects based on forces applied to them.To find the net force, consider all individual forces:
  • Add forces acting in the same direction and subtract those in opposite directions.
The formula is straightforward:
  • \( F_{\text{net}} = F_{\text{forward}} - F_{\text{resistive}} \)
This net force is critical because it's used in the work-energy theorem to calculate work done, determining changes in kinetic energy and thereby affecting an object's speed. Remember, only the net force changes an object's energy, as it's the force that actually causes acceleration.

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

A child of mass \(m\) starts from rest and slides without friction from a height \(h\) along a curved waterslide (Fig. P5.46). She is launched from a height \(h / 5\) into the pool. (a) Is mechanical energy conserved? Why? (b) Give the gravitational potential energy associated with the child and her kinetic energy in terms of \(m g h\) at the following positions: the top of the waterslide, the launching point, and the point where she lands in the pool. (c) Determinc her initial speed \(v_{0}\) at the launch point in terms of \(g\) and \(h\). (d) Determine her maximum airborne height \(y_{\max }\) in terms of \(h, g\), and the horizontal speed at that height, \(v_{0 x}\) (e) Use the \(x\) -component of the answer to part (c) to climinate \(v_{0}\) from the answer to part (d), giving the height \(y_{\text {emax }}\) in terms of \(g, h\), and the launch angle \(\theta\). (f) Would your answers be the same if the waterslide were not frictionless? Explain.

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\) up a \(20^{\circ}\) inclined track. The combined mass of monkey and sled is \(20 \mathrm{~kg}\), and the coefficient of kinetic friction between sled and incline is \(0.20\). How far up the incline do the monkey and sled move?

A child and a sled with a combined mass of \(50.0 \mathrm{~kg}\) slide down a frictionless slope. If the sled starts from rest and has a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) at the bottom, what is the height of the hill?

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by an engine is \(175 \mathrm{hp}\), estimate the total frictional force acting on the car when it is moving at a speed of \(29 \mathrm{~m} / \mathrm{s}\). One horsepower equals \(746 \mathrm{~W}\).

A certain rain cloud at an altitude of \(1.75 \mathrm{~km}\) contains \(3.20 \times 10^{7} \mathrm{~kg}\) of water vapor, How long would it take for a \(2.70-\mathrm{kW}\) pump to raise the same amount of water from Earth's surface to the cloud's position?
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Kinematics Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Solid State Physics

Read Explanation

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation
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.