Problem 33 - University Physics with Modern Physics

Question

A small block with mass 0.0400 kg is moving in the \(xy\)-plane. The net force on the block is described by the potentialenergy function \(U(x, y) = (5.80 \, \mathrm{J/m}^2)x^2 - (3.60 \, \mathrm{J/m}^3)y^3\). What are the magnitude and direction of the acceleration of the block when it is at the point (\(x =\) 0.300 m, \(y =\) 0.600 m)?

Step-by-Step Solution

Answer

The magnitude is \( 5.72 \ \mathrm{m/s}^2\) and the direction is \( 92.5^\circ \) from the x-axis.

Step 1: Identify given quantities

Extract initial velocity, acceleration, time, displacement.

Step 2: Choose kinematic equation

Use v=v0+at, x=v0t+0.5at^2, or v^2=v0^2+2a*dx.

Step 3: Substitute and solve

Plug in values and solve.

Step 4: State the answer

The answer is: The magnitude is \( 5.72 \ \mathrm{m/s}^2\) and the direction is \( 92.5^\circ \) from the x-axis.

Key Concepts

Potential Energy

Potential energy is a concept in physics that refers to the energy stored in an object due to its position or configuration. In classical mechanics, potential energy is often associated with objects under the influence of gravitational, elastic, or electrical forces.

In the example given, the potential energy function, which depends on the "x" and "y" coordinates, signifies the energy stemming from a combination of forces acting on the block. Potential energy functions allow the calculation of force exerted on an object by applying derivatives (more on this shortly). The potential energy function used in this problem contains terms for both the "x" and "y" directions:

• For the "x" direction: \((5.80 \, \mathrm{J/m}^2)x^2\)
  
• For the "y" direction: \(-(3.60 \, \mathrm{J/m}^3)y^3\).

The coefficients and powers of "x" and "y" provide insight into the intensity and nature of the forces in each direction.

Force Components

Understanding force components is crucial for analyzing motion in more than one dimension, like in the given problem. Forces can be resolved into components to simplify calculations and understand how they act. In this scenario, forces in the "x" and "y" directions can be derived from the potential energy function using its gradient.

The gradient is a vector that indicates the direction and rate of the fastest increase of a function. To find the force components related to a potential energy function, we use negative partial derivatives. The equations are:

• For the "x" component of force \(F_x\): \(F_x = -\frac{\partial U}{\partial x}\) = \( -2(5.80 \, \mathrm{J/m}^2)x = -11.6x \, \mathrm{J/m}^2 \)


• For the "y" component of force \(F_y\): \( F_y = -\frac{\partial U}{\partial y}\) = \( 3(3.60 \, \mathrm{J/m}^3)y^2 = 10.8y^2 \, \mathrm{J/m}^2 \)

This breakdown is critical for calculating the precise effect of the net force on the block's motion.

Acceleration Calculation

Once the force components are determined, calculating acceleration becomes a straightforward task in classical mechanics. According to Newton's second law of motion, the net force acting on an object is equal to the mass of the object times its acceleration. This can be expressed as:

\[ F = ma \]
Where "F" is the total force, "m" is the mass, and "a" is the acceleration. If we have components for force, acceleration can be likewise decomposed into components:

• \( a_x = \frac{F_x}{m} \)


• \( a_y = \frac{F_y}{m} \)

Substituting values gives us the "x" and "y" components of acceleration. Knowing the location of the block (\(x = 0.300 \, \mathrm{m}\), \(y = 0.600 \, \mathrm{m}\)), these forces can be evaluated as specific numbers, which are then divided by the block’s mass of \(0.0400 \, \mathrm{kg}\)to find each acceleration component. From there, the Pythagorean theorem can be utilized to find the magnitude of the total acceleration.

Gradient of a Function

The gradient of a function is a powerful mathematical tool for understanding how a function changes. In more basic terms, it is a vector composed of the function’s partial derivatives with respect to each variable. It points in the direction of fastest increase of the function and its magnitude represents the rate of increase.

When applied to a potential energy function, the gradient helps identify force directions. This is crucial because in mechanics, force and potential energy are closely related by the negative gradient:

• \( abla U(x,y) = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) \)

By calculating the gradient components, we can glean a lot of information quickly about how potential energy affects the dynamics of a system, such as inducing or changing motion. Consider how the gradient directly leads to force components in classical mechanics, making it vital for comprehensive analysis and problem-solving.