/*! 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 5 Which of the following is not a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following is not a valid potential energy function for the spring force \(F=-k x ?\) a) \(\left(\frac{1}{2}\right) k x^{2}\) b) \(\left(\frac{1}{2}\right) k x^{2}+10 \mathrm{~J}\) c) \(\left(\frac{1}{2}\right) k x^{2}-10 \mathrm{~J}\) d) \(-\left(\frac{1}{2}\right) k x^{2}\) e) None of the above is valid.

Short Answer

Expert verified
(a) \((\frac{1}{2}) k x^{2}\) (b) \((\frac{1}{2}) k x^{2} + 10\) (c) \((\frac{1}{2}) k x^{2} - 10\) (d) \(-(\frac{1}{2}) k x^{2}\) Answer: (d) \(-(\frac{1}{2}) k x^{2}\)

Step by step solution

01

Identify the valid potential energy functions

We already know that the standard potential energy function for spring force is \(U(x)=\left(\frac{1}{2}\right) k x^{2}\). Adding or subtracting a constant value, in this case \(10 \mathrm{~J}\), does not change the fact that the potential energy function is still valid. So, we can say that options a), b), and c) are valid potential energy functions.
02

Examine the remaining option

Now, we only have to check option d) to determine whether it is a valid potential energy function for the spring force or not. The given function is \(U(x)=-\left(\frac{1}{2}\right) k x^{2}\). Comparing it to the standard potential energy function, we can see that it has a negative sign in front of the expression. This means that the potential energy will be negative, which is not possible for a spring system, as the potential energy should always be non-negative.
03

Determine the answer

Based on our analysis, we can conclude that option d) is not a valid potential energy function for the spring force. Therefore, the correct answer is: (d) \(-\left(\frac{1}{2}\right) k x^{2}\)

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.

Spring Force
The spring force is a fascinating and important concept in physics. It's the force exerted by a spring when it is compressed or stretched. This force is crucial in understanding how springs work, whether in a mattress, a car suspension system, or even in a simple toy. A spring follows Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. This can be expressed as:\[ F = -kx \]- **F** is the force exerted by the spring (in newtons).- **k** is the spring constant, which measures the stiffness of the spring (in newtons per meter).- **x** is the displacement of the spring from its equilibrium position (in meters). The negative sign indicates that the spring force is a restoring force, meaning it acts in the opposite direction to the displacement to bring the spring back to its original position. Understanding spring force is critical for solving problems related to oscillations, like those involving pendulums or the natural frequency of objects.
Conservative Forces
Conservative forces are forces where the work done on an object is path-independent and only depends on the initial and final positions. In simpler terms, this means the total mechanical energy (kinetic + potential) does not change as an object moves between two points under the influence of a conservative force. Examples of conservative forces include:
  • Gravitational force
  • Electrostatic force
  • Spring force
For springs, the work done by the spring force can be completely described by the change in potential energy. Potential energy functions, such as the standard spring potential energy \( U(x) = \left(\frac{1}{2}\right) k x^2 \), represent stored energy due to the position of an object in the field of a conservative force. Thus, springs are ideal systems for illustrating the principles of conservative forces.
Mechanical Energy
Mechanical energy is a pivotal concept that combines kinetic energy and potential energy. This combination helps us understand the behavior of objects in motion and under the influence of various forces. Mechanical energy can be expressed as:\[ E_{mech} = K + U \]- **\(E_{mech}\)** is the total mechanical energy.- **K** is the kinetic energy.- **U** is the potential energy.In a closed system without external forces (such as friction), the mechanical energy remains constant. This is known as the conservation of mechanical energy and is an essential principle in physics.For a spring system, the potential energy is stored as the spring is either compressed or stretched:\[ U(x) = \left(\frac{1}{2}\right) k x^2 \]While the kinetic energy is derived from the motion of the mass attached to the spring, the conservation of mechanical energy within such a system is a foundational aspect of solving physics problems related to oscillations and dynamics. By understanding how mechanical energy is conserved, students can predict the future motion of objects in these systems effectively.

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 basketball of mass \(0.624 \mathrm{~kg}\) is shot from a vertical height of \(1.2 \mathrm{~m}\) and at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). After reaching its maximum height, the ball moves into the hoop on its downward path, at \(3.05 \mathrm{~m}\) above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

In 1896 in Waco, Texas, William George Crush, owner of the K-T (or "Katy") Railroad, parked two locomotives at opposite ends of a 6.4 -km-long track, fired them up, tied their throttles open, and then allowed them to crash head- on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming that each locomotive weighed \(1.2 \cdot 10^{6} \mathrm{~N}\) and its acceleration along the track was a constant \(0.26 \mathrm{~m} / \mathrm{s}^{2},\) what was the total kinetic energy of the two locomotives just before the collision?

A 1.00 -kg block initially at rest at the top of a 4.00 -m incline with a slope of \(45.0^{\circ}\) begins to slide down the incline. The upper half of the incline is frictionless, while the lower half is rough, with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.300\). a) How fast is the block moving midway along the incline, before entering the rough section? b) How fast is the block moving at the bottom of the incline?

The molecular bonding in a diatomic molecule such as the nitrogen \(\left(\mathrm{N}_{2}\right)\) molecule can be modeled by the Lennard Jones potential, which has the form $$ U(x)=4 U_{0}\left(\left(\frac{x_{0}}{x}\right)^{12}-\left(\frac{x_{0}}{x}\right)^{6}\right) $$ where \(x\) is the separation distance between the two nuclei and \(x_{0}\), and \(U_{0}\) are constants. Determine, in terms of these constants, the following: a) the corresponding force function; b) the equilibrium separation \(x_{0}\), which is the value of \(x\) for which the two atoms experience zero force from each other; and c) the nature of the interaction (repulsive or attractive) for separations larger and smaller than \(x_{0}\).

A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

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.