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Chapter 7: Problem 33

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

Short Answer

Expert verified
The magnitude of the acceleration of the block at the point \(x=0.300 \mathrm{~m}, y=0.600 \mathrm{~m}\) is \(130 \mathrm{m/s}^{2}\) and the direction is \(48.0\degree\) with respect to the x-axis.

Step by step solution

01

Determining the force from the potential function

Firstly, the force in the x and y directions should be calculated using the potential-energy function. The force derived from a potential function \(-U(x,y)\) is directed along the negative gradient of this potential. \ Generally, with \(U(x, y)\), the force \(F\) is found by taking the negative of the derivative of \(U\) with respect to \(x\) and \(y\). Therefore, the force components \(F_x\) and \(F_y\) will be given by: \(F_x = - \frac{dU}{dx}\) and \(F_y = - \frac{dU}{dy}\) respectively.\ After differentiating \(U(x, y)=\left(5.80 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}-\left(3.60 \mathrm{~J} / \mathrm{m}^{3}\right) y^{3}\) with respect to \(x\) and \(y\), we have: \(F_x = -2 (5.80 \mathrm{~J} / \mathrm{m}^{2}) x\) and \(F_y = 3 (3.60 \mathrm{~J} / \mathrm{m}^{3}) y^{2}\).
02

Calculating the force at the given point

Next, the force at the point \(x=0.300 \mathrm{~m}, y=0.600 \mathrm{~m}\) needs to be calculated. This gives us \(F_x = -2 (5.80 \mathrm{~J} / \mathrm{m}^{2}) (0.300 \mathrm{~m}) = -3.48 \mathrm{N}\) and \(F_y = 3 (3.60 \mathrm{~J} / \mathrm{m}^{3}) (0.600 \mathrm{~m})^{2} = -3.888 \mathrm{N}\).
03

Calculating the acceleration at the given point

Since force and acceleration are related through Newton's second law \(F=ma\), we can find the acceleration \(a\) by dividing the force by the mass \(m\) of the block. This gives us: \(a_x = F_x/ m = -3.48 \mathrm{~N} / 0.0400 \mathrm{~kg} = -87.0 \mathrm{m/s}^{2}\) and \(a_y = F_y/ m = -3.888 \mathrm{~N} / 0.0400 \mathrm{~kg} = -97.2 \mathrm{m/s}^{2}\). These are the accelerations in the x and y directions respectively.
04

Calculating the acceleration magnitude and direction

The magnitude of the acceleration is found by taking the square root of the sum of the squares of \(a_x\) and \(a_y\), which results in \(a = \sqrt{(-87.0 \mathrm{m/s}^{2})^{2}+(-97.2 \mathrm{m/s}^{2})^{2}} = 130 \mathrm{m/s}^{2}\). For the direction, we take the arctangent of the ratio of \(a_y\) to \(a_x\) and convert the result to degrees. This results in \(\theta = \arctan(|a_y / a_x|) = \arctan(|-97.2 / -87.0|) = 48.0\degree\). This is the angle between the acceleration vector and the x-axis.

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Key Concepts

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

Potential Energy
Potential energy, in a physics context, refers to the stored energy in an object due to its position or configuration. In this problem, the potential energy function depends on the position of a block in the xy-plane. The function provided is:
  • \( U(x, y) = (5.80 \, \mathrm{J}/\mathrm{m}^2) x^2 - (3.60 \, \mathrm{J}/\mathrm{m}^3) y^3 \)
This means the energy varies with both its 'x' and 'y' coordinates. Potential energy functions like this help us understand how forces act on a block. When a block is at a certain point in space, this function tells us how energy changes due to its position. From this, you can calculate forces that push or pull depending on where the block is located. That's why knowing potential energy is crucial to predict the motion of a block based on its position.
Newton's Second Law
One of the core principles in physics is Newton's Second Law of Motion. This law connects the concepts of force, mass, and acceleration in a simple yet powerful equation:
  • \( F = ma \)
Where:
  • \( F \) is the net force acting on an object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration
In this problem, once the forces on the block are found from the potential energy function, they can be used to find the acceleration. The mass of the block is given as
  • \( 0.0400 \, \mathrm{kg} \)
Newton's Second Law helps us convert the calculated forces into the block's acceleration, allowing us to predict how quickly the block speeds up or slows down while moving across the xy-plane based on the forces computed from the potential energy.
Force Calculation
To determine the force from a given potential energy function, you need to evaluate the negative gradient of the potential energy. The gradient is simply the vector of partial derivatives of the function. For the potential energy function:
  • \( U(x, y) = (5.80 \, \mathrm{J}/\mathrm{m}^2) x^2 - (3.60 \, \mathrm{J}/\mathrm{m}^3) y^3 \)
The force components in the x and y directions are:
  • \( F_x = - \frac{dU}{dx} = -2 (5.80 \, \mathrm{J}/\mathrm{m}^2) x \)
  • \( F_y = - \frac{dU}{dy} = 3 (3.60 \, \mathrm{J}/\mathrm{m}^3) y^2 \)
Calculating these derivatives gives you the respective force at any point
  • \( (x, y) \)
Understanding how to calculate forces from a potential energy function is important as it shows which direction and how strongly a force is acting on an object at different positions.
Acceleration
Acceleration tells us how quickly an object speeds up or slows down. Using the force calculated from the potential energy function and Newton's Second Law, we can find the acceleration components. The formula
  • \( a = \frac{F}{m} \)
Lets us find acceleration when force \( F \) and mass \( m \) are known. In this case:
  • \( a_x = \frac{F_x}{0.0400 \, \mathrm{kg}} \)
  • \( a_y = \frac{F_y}{0.0400 \, \mathrm{kg}} \)
From the given forces:
  • \( F_x = -3.48 \, \mathrm{N} \)
  • \( F_y = -3.888 \, \mathrm{N} \)
Thus, acceleration in the x and y directions are
  • \( a_x = -87.0 \, \mathrm{m/s}^2 \)
  • \( a_y = -97.2 \, \mathrm{m/s}^2 \)
Lastly, the magnitude of acceleration can be found using:
  • \( a = \sqrt{a_x^2 + a_y^2} \)
Which gives the overall acceleration magnitude of the block at the specified point. Understanding how acceleration is calculated helps you understand motion dynamics in various directions.

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Most popular questions from this chapter

A cutting tool under microprocessor control has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=-\operatorname{axy}^{2} \hat{\jmath},\) a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. For \(\alpha=2.50 \mathrm{~N} / \mathrm{m}^{3}\), consider the displacement of the tool from the origin to the point \((x=3.00 \mathrm{~m}, y=3.00 \mathrm{~m}) .\) (a) Calculate the work done on the tool by \(\boldsymbol{F}\) if this displacement is along the straight line \(y=x\) that connects these two points. (b) Calculate the work done on the tool by \(F\) if the tool is first moved out along the \(x\) -axis to the point \((x=3.00 \mathrm{~m}, y=0)\) and then moved parallel to the \(y\) -axis to the point \((x=3.00 \mathrm{~m}, y=3.00 \mathrm{~m}) .(\mathrm{c})\) Compare the work done by \(\vec{F}\) along these two paths. Is \(F\) conservative or nonconservative? Explain.

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

CALC The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

\(\mathrm{A} 15.0 \mathrm{~kg}\) stone slides down \begin{tabular}{l} a snow- covered hill (Fig. \(\mathbf{P 7 . 4 5}\) ), \\ \hline \end{tabular} leaving point \(A\) at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). There is no friction on the hill between points \(A\) and \(B,\) but there is friction on the level ground at the bottom of the hill, between \(B\) and the wall. After entering the rough horizontal region, the stone travels \(100 \mathrm{~m}\) and then runs into a very long. light spring with force constant \(2.00 \mathrm{~N} / \mathrm{m}\). The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and \(0.80,\) respectively. (a) What is the speed of the stone when it reaches point \(B ?\) (b) How far will the stone compress the spring? (c) Will the stone move again after it has been stopped by the spring?

CALC In an experiment, one of the forces exerted on a proton is \(\overrightarrow{\boldsymbol{F}}=-\alpha x^{2} \hat{\imath},\) where \(\alpha=12 \mathrm{~N} / \mathrm{m}^{2}\). (a) How much work does \(\boldsymbol{F}\) do when the proton moves along the straight-line path from the point \((0.10 \mathrm{~m}, 0)\) to the point \((0.10 \mathrm{~m}, 0.40 \mathrm{~m}) ?\) (b) Along the straight-line path from the point \((0.10 \mathrm{~m}, 0)\) to the point \((0.30 \mathrm{~m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{~m}, 0)\) to the point \((0.10 \mathrm{~m}, 0) ?\) (d) Is the force \(\vec{F}\) conservative? Explain. If \(\boldsymbol{F}\) is conservative, what is the potential-energy function for it? Let \(U=0\) when \(x=0\).
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