Potential energy Function

In summary, the potential energy function U(x) is given by U(x) = a/x^2 + bx. To find where the force described is 0, in terms of ab and b, take the derivative and set it to 0. For part b, where a = 10.0 J/m^2 and b = 2.00 J/m, the total energy of 20.0 J can be solved for by setting U(x) equal to Umax and solving for x. This can be done by dividing the equation by (x-1) and solving the resulting quadratic equation.
  • #1
veronicak5678
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0

Homework Statement



A potential energy function is given by U (x) = a/x^2 + bx.
a- Find where the force described is 0, in terms of ab and b.
b- Suppose a = 10.0 J/m^2 and b = 2.00 J/m. if an object has a total energy of 20.0 J, for what values of x would it be limited?

Homework Equations





The Attempt at a Solution



I don't understand this. Should I just take the derivative and set it to 0? Then, for part b, set the entire function equal to 20?
 
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  • #2
(a) Yes.

(b) You're close, and would likely receive most of the available credit by doing that.
It says total energy is 20 J, whereas the U(x) expression is for potential energy. They are not the same thing, so setting them equal is not quite right.
 
  • #3
Thanks for your help, but I ams till having trouble. For part a, I came up with
x = - ( 2a/b) ^ (1/3).

I think that is right, but I have been working on part b for almost half an hour, and I cannot solve it. I have 10mx^2 - x^3 -mx^2K = 0 where K is the unkown kinetic energy. I don't know how to solve this, or if it is only so difficult because I made a mistake somewhere.
 
  • #4
This may help for part a:

dU=-f*ds=0

For part b You have the right idea:

U(x)=Umax and solve for x.
 
  • #5
I'm sorry, I don't understand... my answer for part a is wrong? I thought I should just take the derivative, set it to 0, and solve for x. And for part b, how do I solve it? Do I need to use the quadratic equation or something? I feel like I did it all wrong.
 
  • #6
-2*x^3-20*x^2+10=0
x^3+10*x^2-5=0
solve for x
hint divide by (x-1)
 
Last edited:

1. What is potential energy function?

Potential energy function is a mathematical representation of the energy possessed by an object due to its position or configuration in a given system. It is a scalar function that depends on the position of the object within the system.

2. What are the different types of potential energy function?

There are several types of potential energy function, including gravitational potential energy, elastic potential energy, electric potential energy, and chemical potential energy. Each type of potential energy depends on different factors such as mass, distance, and electric charge.

3. How is potential energy function related to kinetic energy?

Potential energy function and kinetic energy are two forms of energy that are related to each other through the principle of conservation of energy. When an object moves from a higher potential energy state to a lower one, its potential energy decreases while its kinetic energy increases. Similarly, when an object moves from a lower potential energy state to a higher one, its potential energy increases while its kinetic energy decreases.

4. What is the significance of potential energy function in physics?

Potential energy function is a fundamental concept in physics, as it helps us understand the behavior of objects and systems in terms of energy. It allows us to predict the motion and interactions of objects based on their positions and configurations within a system. Potential energy function is also crucial in many practical applications, such as in designing structures and machines.

5. How is potential energy function measured or calculated?

Potential energy function is typically measured or calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object from a reference point. However, the exact method of measurement or calculation may vary depending on the specific type of potential energy and the system in question.

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