Problem 74 - University Physics with Modern Physics

Question

A small object with mass \(m =\) 0.0900 kg moves along the \(+x\)-axis. The only force on the object is a conservative force that has the potential-energy function \(U(x) = -ax^2 + bx^3\), where \(\alpha =\) 2.00 J/m\(^2\) and \(\beta =\) 0.300 J/m\(^3\). The object is released from rest at small \(x\). When the object is at \(x =\) 4.00 m, what are its (a) speed and (b) acceleration (magnitude and direction)? (c) What is the maximum value of \(x\) reached by the object during its motion?

Step-by-Step Solution

Answer

(a) Speed at 4.00 m is 16.8 m/s. (b) Acceleration at 4.00 m is 17.8 m/s² in the +x-direction. (c) Maximum value of x is 6.67 m.

Step 1: Understand the Given Potential Energy Function

The potential energy function is given by \(U(x) = -\alpha x^2 + \beta x^3\), where \(\alpha = 2.00\) J/m\(^2\) and \(\beta = 0.300\) J/m\(^3\). This function describes how potential energy varies with position \(x\).

Step 2: Find the Force as a Function of Position

The force corresponding to the potential energy \(U(x)\) is given by the negative gradient of \(U(x)\). Therefore, \(F(x) = -\frac{dU(x)}{dx}\). First, find \(\frac{dU(x)}{dx} = -2\alpha x + 3\beta x^2\), then \(F(x) = 2\alpha x - 3\beta x^2\).

Step 3: Calculate the Initial Total Energy

Since the object is released from rest, its initial kinetic energy is zero, and total energy is purely potential, \(E_{initial} = U(x_{initial})\). Let \(x_{initial} = 0\) (released from small \(x\)). Therefore, \(E_{initial} = U(0) = 0\).

Step 4: Calculate the Total Energy at x = 4.00 m

The total mechanical energy \(E = K + U\), where \(K\) is kinetic energy. At \(x = 4.00\, \text{m}\), total energy is \(E = K + U(4.00)\). Calculate \(U(4.00) = -2.00 \times (4.00)^2 + 0.300 \times (4.00)^3 = -32.0 + 19.2 = -12.8\, \text{J}\). Since initial energy was zero, \(E = 0\), thus \(K = -U(4) = 12.8\, \text{J}\).

Step 5: Calculate Speed at x = 4.00 m

Kinetic energy is \(K = \frac{1}{2}mv^2\). We have \(12.8 = \frac{1}{2} \times 0.0900 \times v^2\). Solve for \(v\) to get \(v = \sqrt{\frac{12.8 \times 2}{0.0900}} = 16.8\, \text{m/s}\).

Step 6: Find the Acceleration at x = 4.00 m

Use Newton's second law, \(F = ma\). From Step 2, \(F(4.00) = 2 \times 2.00 \times 4.00 - 3 \times 0.300 \times 4.00^2 = 16.0 - 14.4 = 1.6\, \text{N}\). Therefore, \(a = \frac{1.6}{0.0900} = 17.8\, \text{m/s}^2\). The force is positive, so the acceleration is in the \(+x\)-direction.

Step 7: Determine Maximum Value of x

The maximum value of \(x\) occurs when kinetic energy is zero and potential energy is maximum. The energy was initially zero at \(x = 0\), and since energy is conserved, the potential energy at maximum \(x\) equals zero. Solve \(-2.00 x^2 + 0.300 x^3 = 0\). This gives \(x ( -2 + 0.300x) = 0\). Either \(x = 0\) or \(x = \frac{20}{3.0} = 6.67\) m. Thus, the maximum \(x\) is 6.67 m.

Key Concepts

Potential Energy Function

In physics, the potential energy function describes how potential energy varies within a system based on position. It's an essential part of understanding conservative forces, which are linked to potential energy. For this exercise, the potential energy function is given by:
\[ U(x) = -\alpha x^2 + \beta x^3 \]where \( \alpha = 2.00 \) J/m\(^2\) and \( \beta = 0.300 \) J/m\(^3\).
This formula allows us to predict the potential energy at different positions \( x \). Calculating potential energy is crucial as it helps indicate the system's behavior without needing to consider kinetic energy immediately. By analyzing how these potential energies vary, one can also predict the movement of an object under conservative forces.

Mechanical Energy Conservation

Mechanical energy is the sum of potential and kinetic energy in a system. In a system involving only conservative forces, like the one in this exercise, mechanical energy is conserved.

• Mechanical energy conservation means that the total energy of the system remains constant over time.
• This principle helps us calculate unknown properties, such as velocity and position, as energy shifts between potential and kinetic forms.

For the problem at hand, we can conclude that the total mechanical energy at the start is zero. As the object moved to \( x = 4.00 \) m, the potential energy changes, but mechanical energy conservation allows us to find the missing kinetic energy, maintaining a total energy balance.

Kinetic Energy Calculation

Kinetic energy depends on the mass and velocity of an object. Given by the equation:
\[ K = \frac{1}{2} m v^2 \]we can solve for unknown variables like speed if we know the kinetic energy.
From the exercise, once at \( x = 4 \) m, we learned the potential energy, so by conservation, we discovered the kinetic energy to be \( 12.8 \) J. Consequently, solving:
\[ 12.8 = \frac{1}{2} \times 0.0900 \times v^2 \]can lead us towards calculating the speed as \( 16.8 \) m/s.
Kinetic energy calculations are vital for understanding the dynamic state of a system, showing how potential energy transforms into motion.

Force and Acceleration Relationship

The relationship between force and acceleration is dictated by Newton's second law, expressed as:
\[ F = ma \]This allows us to find acceleration if the force and mass are known. In our scenario, force is not directly given but can be calculated from the potential energy function:

• The force is the negative derivative of the potential energy function, \( F(x) = -\frac{dU}{dx} \).
• In this case, \( F(x) = 2\alpha x - 3\beta x^2 \).

Plugging in \( x = 4.00 \) m gives \( F(4) = 1.6 \) N. Using \( F = ma \), the acceleration becomes \( 17.8 \) m/s\(^2\) in the positive x-direction. Understanding this relationship helps us predict how an object will move under varying forces.