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107 The only force acting on a particle is conservative force \(\vec{F}\). If the particle is at point \(A\), the potential energy of the system associated with \(\vec{F}\) and the particle is \(40 \mathrm{~J}\). If the particle moves from point \(A\) to point \(B\), the work done on the particle by \(\vec{F}\) is \(+25 \mathrm{~J}\). What is the potential energy of the system with the particle at \(B\) ?

Short Answer

Expert verified
The potential energy at point B is 15 J.

Step by step solution

01

Understand the Concept of Conservative Force

A conservative force is a type of force for which the work done only depends on the initial and final positions of the object, independent of the path taken. The conservation of mechanical energy equation can be used: \[ U_A - W = U_B \] where \( U_A \) is the initial potential energy, \( W \) is the work done, and \( U_B \) is the final potential energy.
02

Identify Given Values

From the problem, the potential energy at point \( A \) (\( U_A \)) is given as \( 40 \text{ J} \). The work done by the force as the particle moves from \( A \) to \( B \) is \(+25 \text{ J}\). We need to find the potential energy at point \( B \) (\( U_B \)).
03

Set Up the Equation

Using the equation for conservation of mechanical energy \( U_A - W = U_B \), substitute the given values: \[ 40 \text{ J} - 25 \text{ J} = U_B \] This equation allows us to find the potential energy at point \( B \).
04

Solve for Potential Energy at Point B

Calculate \( U_B \) by subtracting the work done from the initial potential energy: \[ U_B = 40 \text{ J} - 25 \text{ J} = 15 \text{ J} \] Thus, the potential energy at point \( B \) is \( 15 \text{ J} \).

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Key Concepts

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

Potential Energy
Potential energy is a vital concept in physics, particularly when dealing with forces. It represents the energy stored in an object due to its position or configuration. Let's consider gravitational potential energy, a common type often encountered. When you lift a book off the ground, it gains potential energy due to its height above the earth. This energy is stored in the book because of the gravitational force acting upon it, which is a type of conservative force.
When discussing potential energy with conservative forces, it's crucial to remember that such forces only depend on the object's starting and ending points. The path taken does not change the potential energy involved but only these key positions matter.
In the context of our example, the initial potential energy at point A is given as 40 Joules. As the particle moves to point B under a conservative force, calculations of potential energy consider this initial value and any work done along the way.
Work-Energy Principle
The work-energy principle is a foundational concept in physics. It states that the work done by a force on an object is equal to the change in its kinetic energy. When dealing with conservative forces, this principle can simplify to show relationships between work done and changes in potential energy.
The formula for the work-energy principle can be written as:
  • \[ W = ext{Change in Kinetic Energy (K.E.)} \]
Where \( W \) is the work done, which acts as the bridge between potential and kinetic energies. This principle is especially helpful in solving problems involving moving particles under conservative forces.
In our original exercise, the work done as the particle moves from point A to point B was +25 Joules. By utilizing the work-energy principle, we can connect this work with the change in potential energy from point A to point B.
Mechanical Energy Conservation
Mechanical energy conservation is a principle that states that the total mechanical energy in a system remains constant if only conservative forces are acting upon it. Mechanical energy is the sum of potential energy and kinetic energy. This principle is a direct result of the characteristics of conservative forces, where energy is neither lost nor gained but simply transformed from one form to another.
The equation used to express this conservation is:
  • \[ U_A + K_A = U_B + K_B \]
Where \( U \) represents potential energy and \( K \) represents kinetic energy at points A and B, respectively.
In the problem, although we were specifically dealing with potential energy, understanding that mechanical energy was conserved allows us to conclude that kinetic energy would also change in such a way to keep the total sum constant, if kinetic energies were given and needed to be considered. This is why the potential energy could decrease by the 25 J of work done (from 40 J to 15 J), without "losing" any total mechanical energy.

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Most popular questions from this chapter

Approximately \(5.5 \times 10^{6} \mathrm{~kg}\) of water falls \(50 \mathrm{~m}\) over Niagara Falls each second. (a) What is the decrease in the gravitational potential energy of the water-Earth system each second? (b) If all this energy could be converted to electrical energy (it cannot be), at what rate would electrical energy be supplied? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(\left.1000 \mathrm{~kg} .\right)(\mathrm{c})\) If the electrical energy were sold at 1 cent \(/ \mathrm{kW} \cdot \mathrm{h}\), what would be the yearly income?

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\), and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N}\), the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\), with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\).

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

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