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Chapter 6: Problem 36

The potential energy of a certain particle is given by \(U=10 x^{2}+35 z^{3}\). Find the force vector exerted on the particle.

Short Answer

Expert verified
Answer: The force vector exerted on the particle is \(F = \langle -20x, -105z^2 \rangle\).

Step by step solution

01

Calculate the partial derivative of U with respect to x

To find the partial derivative of U with respect to x, we differentiate U with respect to x, treating z as a constant. \[\frac{\partial U}{\partial x} = 20x\]
02

Calculate the partial derivative of U with respect to z

To find the partial derivative of U with respect to z, we differentiate U with respect to z, treating x as a constant. \[\frac{\partial U}{\partial z} = 105z^2\]
03

Represent the partial derivatives as a force vector

We will represent the force vector (F) as \[F = \langle F_x, F_z \rangle\]
04

Calculate the components of the force vector

To calculate the components of the force vector, we apply the negative gradient of the potential energy function. So, we will have \[F_x = -\frac{\partial U}{\partial x} = -20x\] and \[F_z = -\frac{\partial U}{\partial z} = -105z^2\]
05

Write the force vector

Now that we have the components of the force vector, we can write it as: \[F = \langle -20x, -105z^2 \rangle\] Therefore, the force vector exerted on the particle is \(F = \langle -20x, -105z^2 \rangle\).

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Most popular questions from this chapter

The molecular bonding in a diatomic molecule such as the nitrogen \(\left(\mathrm{N}_{2}\right)\) molecule can be modeled by the Lennard Jones potential, which has the form $$ U(x)=4 U_{0}\left(\left(\frac{x_{0}}{x}\right)^{12}-\left(\frac{x_{0}}{x}\right)^{6}\right) $$ where \(x\) is the separation distance between the two nuclei and \(x_{0}\), and \(U_{0}\) are constants. Determine, in terms of these constants, the following: a) the corresponding force function; b) the equilibrium separation \(x_{0}\), which is the value of \(x\) for which the two atoms experience zero force from each other; and c) the nature of the interaction (repulsive or attractive) for separations larger and smaller than \(x_{0}\).

A ball of mass \(1.84 \mathrm{~kg}\) is dropped from a height \(y_{1}=$$1.49 \mathrm{~m}\) and then bounces back up to a height of \(y_{2}=0.87 \mathrm{~m}\) How much mechanical energy is lost in the bounce? The effect of air resistance has been experimentally found to be negligible in this case, and you can ignore it.

A 20.0 -kg child is on a swing attached to ropes that are \(L=1.50 \mathrm{~m}\) long. Take the zero of the gravitational potential energy to be at the position of the child when the ropes are horizontal. a) Determine the child's gravitational potential energy when the child is at the lowest point of the circular trajectory. b) Determine the child's gravitational potential energy when the ropes make an angle of \(45.0^{\circ}\) relative to the vertical. c) Based on these results, which position has the higher potential energy?

A girl of mass \(49.0 \mathrm{~kg}\) is on a swing, which has a mass of \(1.0 \mathrm{~kg} .\) Suppose you pull her back until her center of mass is \(2.0 \mathrm{~m}\) above the ground. Then you let her \(\mathrm{go},\) and she swings out and returns to the same point. Are all forces acting on the girl and swing conservative?

One end of a rubber band is tied down and you pull on the other end to trace a complicated closed trajectory. If you were to measure the elastic force \(F\) at every point and took its scalar product with the local displacements, \(\vec{F} \cdot \Delta \vec{r},\) and then summed all of these, what would you get?
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