/*! 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 33 A small block with mass 0.0400 k... [FREE SOLUTION] | 91Ó°ÊÓ

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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)?

Short Answer

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

Step by step solution

01

Calculate Force Components

The force exerted by a potential energy function can be calculated using the negative gradient of that function. In two dimensions, this means \( F_x = -\frac{\partial U}{\partial x} \) and \( F_y = -\frac{\partial U}{\partial y} \). Therefore: \[ F_x = -\frac{\partial}{\partial x}((5.80 \ \mathrm{J/m}^2)x^2 - (3.60 \ \mathrm{J/m}^3)y^3) = -2(5.80 \ \mathrm{J/m}^2)x = -11.6x \, \mathrm{J/m}^2 \], and \[ F_y = -\frac{\partial}{\partial y}((5.80 \ \mathrm{J/m}^2)x^2 - (3.60 \ \mathrm{J/m}^3)y^3) = 3(3.60 \ \mathrm{J/m}^3)y^2 = 10.8y^2 \, \mathrm{J/m}^3 \].

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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 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.

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Most popular questions from this chapter

A ball is thrown upward with an initial velocity of 15 m/s at an angle of 60.0\(^\circ\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

A spring of negligible mass has force constant \(k =\) 1600 N/m. (a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20-kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

You are designing an amusement park ride. A cart with two riders moves horizontally with speed \(v = 6.00\) m/s. You assume that the total mass of cart plus riders is 300 kg. The cart hits a light spring that is attached to a wall, momentarily comes to rest as the spring is compressed, and then regains speed as it moves back in the opposite direction. For the ride to be thrilling but safe, the maximum acceleration of the cart during this motion should be 3.00\(g\). Ignore friction. What is (a) the required force constant of the spring, (b) the maximum distance the spring will be compressed?

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with \(y\) positive upward, and (b) at the top of the ramp with y positive upward. (c) Why didn't the normal force enter into your solution?

A 60.0-kg skier starts from rest at the top of a ski slope 65.0 m high. (a) If friction forces do \(-\)10.5 kJ of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow where \(\mu_k\) = 0.20. If the patch is 82.0 m wide and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch? (c) The skier hits a snowdrift and penetrates 2.5 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

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