/*! 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 68 A net force along the \(x\)-axis... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A net force along the \(x\)-axis that has \(x\)-component \(F{_x}= -12.0\mathrm{N} +(0.300\mathrm{N/m{^2}})x{^2}\) is applied to a 5.00-kg object that is initially at the origin and moving in the \(-x\)-direction with a speed of 6.00 m/s. What is the speed of the object when it reaches the point \(x = 5.00\) m?

Short Answer

Expert verified
The speed of the object at \(x = 5.00\) m is approximately 4.12 m/s.

Step by step solution

01

Set up the Work-Energy Theorem

The work-energy theorem states that the work done by all the forces acting on an object equals the change in its kinetic energy. Mathematically, \( W = \Delta K \), where \( W \) is the work done, and \( \Delta K \) is the change in kinetic energy, given by \( \frac{1}{2} m (v_f^2 - v_i^2) \). Here, \( m = 5.00 \) kg, and we need to find the final velocity \( v_f \) when the object reaches \( x = 5.00 \) m.
02

Calculate Initial Kinetic Energy

Initially, the object moves at a speed of 6.00 m/s in the \(-x\)-direction. The initial kinetic energy is given by:\[K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (5.00 \, \text{kg}) (6.00 \, \text{m/s})^2 = 90.0 \, \text{J}.\]
03

Calculate Work Done by the Force

The work done by the force as the object moves from \( x = 0 \) to \( x = 5.00 \) m is found by integrating the force over distance:\[W = \int_{0}^{5.00} \left(-12.0 \text{ N} + (0.300 \text{ N/m}^2) x^2\right) dx.\]Integrate term by term:\[W = \left[-12.0x + 0.100x^3\right]_{0}^{5.00}.\]Evaluate this expression:\[W = \left(-12.0 \times 5.00 + 0.100 \times (5.00)^3\right) - \left(-12.0 \times 0 + 0.100 \times 0^3\right) = -60.0 + 12.5 = -47.5 \, \text{J}.\]
04

Calculate Final Kinetic Energy and Speed

Using the work-energy theorem, equate the work done to the change in kinetic energy:\[-47.5 = \frac{1}{2} m (v_f^2 - v_i^2).\]Plug in the known values:\[-47.5 = \frac{1}{2} (5.00) (v_f^2 - 36.0).\]Solving for \( v_f^2 \), we find:\[-47.5 = 2.5 (v_f^2 - 36),\]\[-47.5 = 2.5v_f^2 - 90,\]\[2.5v_f^2 = 42.5,\]\[v_f^2 = 17,\]\[v_f = \sqrt{17} \approx 4.12 \, \text{m/s}.\]

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.

Kinetic Energy
Kinetic energy is the energy of an object due to its motion. It is a crucial concept in physics because it allows us to understand how objects move and interact with forces. The formula for kinetic energy (K) is:
\[K = \frac{1}{2} m v^2\]
where:
  • \(m\) is the mass of the object
  • \(v\) is the velocity of the object
In the given problem, the object initially moves with a velocity of -6.00 m/s. Using the formula, we can calculate its initial kinetic energy:
\[K_i = \frac{1}{2} \times 5.00 \, \text{kg} \times (6.00 \, \text{m/s})^2 = 90.0 \, \text{J}\]
The negative sign indicates that the object is moving in the opposite direction of the positive reference axis. However, kinetic energy is a scalar quantity, so it is always positive, regardless of direction. As the object moves, forces can do work on it, changing its kinetic energy and thereby altering its speed. Understanding how kinetic energy changes in response to forces is essential for solving problems involving motion.
Net Force
The net force is the total force acting on an object when all individual forces are considered. It determines the acceleration and direction of motion of the object according to Newton's second law:
\[F_{\text{net}} = m a\]
In the exercise, the net force along the x-axis is given by its x-component:
\[F_x = -12.0 \, \text{N} + 0.300 \, \text{N/m}^2 \times x^2\]
This force has two parts:
  • A constant force of -12.0 N that acts constantly as the object moves.
  • A variable force that depends on the square of the position \(x^2\) and increases as the object moves away from the origin.
The forces work together to alter the motion of the object. The negative sign in the constant force suggests that it acts in the opposite direction of the object's initial motion. Calculating the net force helps determine how the velocity and kinetic energy of the object change as it moves.
Integration in Physics
Integration is a mathematical technique used in physics to find terms related to area, volume, and other accumulated quantities. When an object moves in response to a force, integration helps in calculating the work done by that force over a distance. In this problem, the work done by the force is calculated using the integral of the force expression over the range from \(x = 0\) to \(x = 5.00\, \text{m}\):
\[W = \int_{0}^{5.00} \left(-12.0 \, \text{N} + 0.300 \, \text{N/m}^2 \, x^2 \right) dx\]
Integration allows us to add up all the tiny values of work done along every small movement from \(x\). For our specific force equation, performing the integration term by term results in:
\[W = \left[-12.0x + 0.100x^3\right]_{0}^{5.00}\]
This process helps identify the total work done as -47.5 J, which represents the change in energy needed to compute the object's final velocity using the work-energy theorem.
Understanding integration is vital for analyzing the effects of varying forces over a path, leading to insights into energy changes and mechanics dynamics.

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 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be \(different\) if there were appreciable air resistance?

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman (1.63 m). The density (mass per unit volume) of blood is \(1.05 \times 10^3 \, \mathrm{kg/m}^3\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 km along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0\(^\circ\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

Three identical 8.50-kg masses are hung by three identical springs (\(\textbf{Fig. E6.35}\)). Each spring has a force constant of 7.80 kN/m and was 12.0 cm long before any masses were attached to it. (a) Draw a free-body diagram of each mass. (b) How long is each spring when hanging as shown? (\(Hint\): First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)

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.