Why Is the Derivative of Potential Energy Considered Force?

[transgalactic]

the potential energy of a body which mass is "m"
and which is moving in one demention is given by:

U(x)=U0 * (a/x + x/a -2)
what is the power which being put on the body

the solution says F=-dU/dx=U0(-1/a + a/x^2)

which means that they are doing a derivative to the original expression.

but i can't understand why the derivative of potential energy is its power.

from high school i know the potential energy is U=m*g*h

and the only derivative that i know is for a(t)=v'(t) v(t)=x'(t)

so their solution is not logic to me at all

?

 

[buffordboy23]

the potential energy of a body which mass is "m"
and which is moving in one demention is given by:

U(x)=U0 * (a/x + x/a -2)
what is the power which being put on the body

the solution says F=-dU/dx=U0(-1/a + a/x^2)

which means that they are doing a derivative to the original expression.

but i can't understand why the derivative of potential energy is its power.

U(x) is a scalar function for potential. F=-dU/dx gives the force on the particle, not the power.

Look at the units of your quantities. A force applied over some distance is the "work" done on the particle; this has units of joules. Now, the work done over some time period is the power; P = dW/dt, which has units of joules per seconds. So, yes, something does seem quite strange if that is what your text is telling you.

from high school i know the potential energy is U=m*g*h

and the only derivative that i know is for a(t)=v'(t) v(t)=x'(t)

so their solution is not logic to me at all

?

Potential energy comes in many different forms. U=mgh is the gravitational potential energy for an object some height above Earth's surface. Another example is U=(1/2)kx^2, which is the potential energy of a mass-spring system. Your function U(x) is just some general function for a potential energy; for this problem, it appears that it is not overly important for you to know what type of potential energy it is, but rather, that by taking the derivative with respect to x, you can determine the force on the mass m at any position x where the potential is defined.

 

[transgalactic]

ok i understand that the force which is applied on
is the derivative of U(x)

but the definition of a derivative is dU/dx

why are they writing F=-dU/dx (why they add minus)

??

 

[buffordboy23]

The minus sign is just a convention.

Consider a particle of mass m confined within the potential U(x) = x^2. The force on the particle is given by F=-dU/dx, right? Thus, F = -2x. Suppose that the particle is located at x=a with no initial kinetic energy. What is the force on the particle and its direction? F=-2a and the force acts to push the particle to the left, and down the "hill" (parabola). If there were no minus sign in front of dU/dx, then the force would act to push the particle to the right and into regions of higher potential. The minus sign convention seems to offer a better description in regards to physical reality.