Why is the Sign of Work and Energy Important?

  • Thread starter Thread starter jegues
  • Start date Start date
  • Tags Tags
    Energy Work
AI Thread Summary
The discussion revolves around the confusion regarding the signs of potential energy and work in the context of electric dipoles. When the dipole moment is aligned with the electric field, the potential energy is zero, but as it rotates, the potential energy becomes negative, leading to confusion about which sign to choose for answers. The negative sign in the potential energy formula indicates that the system is in a lower energy state when aligned with the field. Additionally, while the work done appears negative due to the change in potential energy, the correct interpretation is that the work is positive when considering the direction of energy transfer. Clarifying these concepts helps eliminate sign errors in calculations.
jegues
Messages
1,085
Reaction score
3

Homework Statement



There are two problems attached that illustrate my confusions, but my questions is more general than this specific example.

Homework Equations




The Attempt at a Solution



For the first problem where we are asked for the potential energy,

When the dipole moment is parallel to the electric field, it has 0 potential energy.

U_{i} = - \vec{p} \cdot \vec{E} = 0

When the dipole rotates 60 degrees, the potential energy is not 0.

U_{f} = -\vec{p} \cdot \vec{E} = -pEcos\theta

The potential enegry given from the formula is negative, but there are only positive answers as possibilities.

I know it is obvious to choose the positive one that corresponds to our answer, but sometimes both negative a positive quantities are listed as possible answers, and I am often confused on which one to choose.

Why is that negative sign in our formula for potential energy, what's its purpose?

For problem number 2 where we are asked for the work,

Initially there is 0 potential energy in the system (perpendicular),

U_{i} = 0

but then the dipole rotates such that it is parallel to the direction of the field so,

U_{f} = -pE

Now,

W = \Delta U = U_{f} - U_{i} = -pE - 0 = -pE

It seems as though the work is negative, but the correct answer is that it is positive.

What am I mixing up?

What do I need to think about and be consistent with to eliminate these sign errors?

Please be as clear as possible. A good example to remember how to determine the sign would be excellent.

Thanks again!
 

Attachments

  • UCONF.JPG
    UCONF.JPG
    21.5 KB · Views: 368
  • WORKCONF.JPG
    WORKCONF.JPG
    23.8 KB · Views: 375
Physics news on Phys.org
Bump, still looking for some help on this, tests tomorrow!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top