Chapter 2: Problem 8
A particle of mass \(m\) has the potential energy function \(V(x)=m k|x|\) where \(k\) is a positive constant. What is the force when \(x>0\), and when \(x<0 ?\) Sketch the function \(V\) and describe the motion. If the particle starts from rest at \(x=-a\), find the time it takes to reach \(x=a\).
Short Answer
Expert verified
Answer: \(t = \sqrt{\frac{4a}{k}}\)
Step by step solution
01
Finding the Force
The force acting on the particle is the negative derivative of the potential energy function with respect to x. So, we need to first find the derivative of V(x) with respect to x.
\(V(x) = mk|x|\)
For x > 0, \(V(x) = mkx\). Now we have to differentiate this with respect to x.
\(\frac{dV(x)}{dx} = mk\)
Since the force is the negative derivative of the potential energy function,
\(F(x) = -\frac{dV(x)}{dx} = -mk\) for \(x > 0\)
For x < 0, V(x) = -mkx. Differentiating this with respect to x,
\(\frac{dV(x)}{dx} = -mk\)
The force on the particle in this case is,
\(F(x) = -\frac{dV(x)}{dx} = mk\) for \(x < 0\)
02
Sketching V(x) and Describing Motion
We have the potential energy function \(V(x) = mk|x|\). It is a linear function for both positive and negative x, so we can describe it as a V-shaped function with a minimum at x = 0.
For \(x<0\), the particle has a positive force acting on it, which means it will be pushed towards the positive x direction. On the other hand, for \(x>0\), the force acting on the particle is negative, so it will be pushed back towards the negative x direction. This means the particle oscillates back and forth between x = -a and x = a.
03
Finding the Time Taken to Reach x = a from x = -a
In this step, we will calculate the time it takes for the particle to move from x = -a to x = a. We will use the equation of motion under constant force for this:
\(s = ut + \frac{1}{2}at^2\)
Since the particle starts from rest at x = -a, the initial velocity u is 0. The displacement s is 2a (from -a to a).
The acceleration a can be calculated using Newton's second law of motion:
\(F = ma\)
Since \(F = mk\), we have:
\(mk = ma\)
This implies that the acceleration a is constant in magnitude and equal to k.
Now we can substitute these values into the equation of motion:
\(2a = 0 + \frac{1}{2}kt^2\)
Simplifying and solving for t gives:
\(t^2 = \frac{4a}{k}\)
Taking the square root of both sides, we get:
\(t = \sqrt{\frac{4a}{k}}\)
So, the time taken for the particle to move from x = -a to x = a is:
\(t = \sqrt{\frac{4a}{k}}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Force and Potential Energy
Understanding the relationship between force and potential energy is crucial to mastering concepts in classical mechanics. Potential energy, often denoted as V(x), is the energy by virtue of an object's position relative to other objects. For a particle with mass m having a potential energy function of the form V(x) = mk|x|, where k is a constant, we can find the associated force.
Force is defined as the negative gradient of the potential energy, which implies a function V(x) that increases with x will result in a force that acts in the negative x-direction, or vice versa. Mathematically, the force F(x) is determined by taking the derivative of the potential energy function with respect to position x and then negating it. For positive x values, the force is negative, indicating that the particle will feel a push back toward the origin. Conversely, for negative x values, the particle experiences a positive force, which will drive the particle toward the positive direction.
This dynamic leads to what is known as a restorative force - the force that always acts to pull the particle back towards a point of equilibrium. In our exercise, this point is at x = 0, where the potential energy is minimum and the force changes direction. This nature of the force has key implications for the motion of the particle, which we explore next.
Force is defined as the negative gradient of the potential energy, which implies a function V(x) that increases with x will result in a force that acts in the negative x-direction, or vice versa. Mathematically, the force F(x) is determined by taking the derivative of the potential energy function with respect to position x and then negating it. For positive x values, the force is negative, indicating that the particle will feel a push back toward the origin. Conversely, for negative x values, the particle experiences a positive force, which will drive the particle toward the positive direction.
This dynamic leads to what is known as a restorative force - the force that always acts to pull the particle back towards a point of equilibrium. In our exercise, this point is at x = 0, where the potential energy is minimum and the force changes direction. This nature of the force has key implications for the motion of the particle, which we explore next.
Equations of Motion
Equations of motion are vital tools for solving classical mechanics problems, particularly when dealing with particles under the influence of constant forces. The basic equations of motion help us describe the kinematics of particles by relating displacement, velocity, acceleration, and time.
For constant acceleration, these are the classical equations we usually refer to:
By substituting known values and rearranging the equations, we can find the unknown variables such as the time it takes for a particle to move from one point to another under a constant force, demonstrating the practical utility of the equations of motion in predicting future states of motion from given conditions.
For constant acceleration, these are the classical equations we usually refer to:
- \( v = u + at \) - Final velocity after time t starting with an initial velocity u
- \( s = ut + \frac{1}{2}at^2 \) - Displacement after time t
- \( v^2 = u^2 + 2as \) - Relating final velocity to displacement
By substituting known values and rearranging the equations, we can find the unknown variables such as the time it takes for a particle to move from one point to another under a constant force, demonstrating the practical utility of the equations of motion in predicting future states of motion from given conditions.
Classical Mechanics Problem Solving
Solving problems in classical mechanics often involves a systematic approach that begins with understanding the forces at play and ends with employing mathematical equations to describe motion. The problem involving a particle acted upon by a force from a V-shaped potential well demonstrates this process.
First, we examine the nature of the force involved by analyzing the potential energy function. Recognizing that force is derived from this potential energy, we calculate the derivative to quantify the force. Next, we consider the motion of the particle under this force, bearing in mind that the direction of force will change with the sign of the position x, leading to oscillatory motion.
Finally, employing one of the key equations of motion, we unravel the duration of such oscillatory journeys. Throughout this process, it is essential to maintain a clear understanding of the physical concepts, such as potential wells and restorative forces, while applying mathematical tools, such as differentiation and the equations of motion, correctly. This combination of physical insight and mathematical prowess results in a robust method for solving a wide range of problems in classical mechanics.
First, we examine the nature of the force involved by analyzing the potential energy function. Recognizing that force is derived from this potential energy, we calculate the derivative to quantify the force. Next, we consider the motion of the particle under this force, bearing in mind that the direction of force will change with the sign of the position x, leading to oscillatory motion.
Finally, employing one of the key equations of motion, we unravel the duration of such oscillatory journeys. Throughout this process, it is essential to maintain a clear understanding of the physical concepts, such as potential wells and restorative forces, while applying mathematical tools, such as differentiation and the equations of motion, correctly. This combination of physical insight and mathematical prowess results in a robust method for solving a wide range of problems in classical mechanics.