Solving Problem with Units: Max Elastic Potential Energy

[jg370]

Homework Statement

In a problem to determine the maximum elastic potential energy of a molecule, I have the following equation:

$$A= \sqrt{\frac{6.626*10^-34 Js}{2\pi}}\times(\frac{1}{(1.14*10^-26kg)(1.85*10^3N/m})^(1/4)$$

Homework Equations

The Attempt at a Solution

The answer is $$4.79*10^-12 m$$

I have tried to coax the definition of the Joule and Newton units to come out with an answer in meter; however I do not see how this can be made to work.

Thank you for your kind assistance

jg370

 

[LowlyPion]

(J*s)^1/2 = kg^1/2*m*s^-1/2

(N/m)^-1/4 = kg^-1/4*m^-1/4*s^1/2*m^1/4

And there hiding in the denominator - the extra kg^-1/4

As they say in poker "Pot's right."

 

[nlightner]

Hello jg370,

Thank you for sharing your problem and attempted solution. It is clear that you are on the right track in trying to solve for the maximum elastic potential energy of a molecule using the given equation. However, there are a few things that need to be addressed in order to arrive at the correct answer.

Firstly, when solving for a physical quantity, it is important to make sure that all units are consistent throughout the equation. In this case, you have correctly identified that the units for energy (Joules) and force (Newtons) need to be converted to the same unit. However, it seems that you may have made a mistake in the conversion.

The equation for elastic potential energy is given by U = 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In your equation, you have the square root of the product of two terms, which suggests that one term is equal to k and the other is equal to x^2. However, the units for k are N/m, not N/m^2. This means that the conversion factor for the force term should be 1.85*10^3 N/m, not 1.85*10^3 N/m^2.

With this correction, the units for the equation will be consistent and you should be able to solve for the maximum elastic potential energy in meters. Keep in mind that the final answer should also be in meters, not meters squared, as the equation for elastic potential energy has units of length.

I hope this helps and good luck with your problem-solving!