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In two dimensions,

F = F x i ^ + F y j ^ = ( U x ) i ^ ( U y ) j ^ .

From this equation, you can see why [link] is the condition for the work to be an exact differential, in terms of the derivatives of the components of the force. In general, a partial derivative notation is used. If a function has many variables in it, the derivative is taken only of the variable the partial derivative specifies. The other variables are held constant. In three dimensions, you add another term for the z -component, and the result is that the force is the negative of the gradient of the potential energy. However, we won’t be looking at three-dimensional examples just yet.

Force due to a quartic potential energy

The potential energy for a particle undergoing one-dimensional motion along the x -axis is

U ( x ) = 1 4 c x 4 ,

where c = 8 N/m 3 . Its total energy at x = 0 is 2 J, and it is not subject to any non-conservative forces. Find (a) the positions where its kinetic energy is zero and (b) the forces at those positions.

Strategy

(a) We can find the positions where K = 0 , so the potential energy equals the total energy of the given system. (b) Using [link] , we can find the force evaluated at the positions found from the previous part, since the mechanical energy is conserved.

Solution

  1. The total energy of the system of 2 J equals the quartic elastic energy as given in the problem,
    2 J = 1 4 ( 8 N/m 3 ) x f 4 .

    Solving for x f results in x f = ± 1 m .
  2. From [link] ,
    F x = d U / d x = c x 3 .

    Thus, evaluating the force at ± 1 m , we get
    F = ( 8 N/m 3 ) ( ± 1 m ) 3 i ^ = ± 8 N i ^ .

    At both positions, the magnitude of the forces is 8 N and the directions are toward the origin, since this is the potential energy for a restoring force.

Significance

Finding the force from the potential energy is mathematically easier than finding the potential energy from the force, because differentiating a function is generally easier than integrating one.

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Check Your Understanding Find the forces on the particle in [link] when its kinetic energy is 1.0 J at x = 0 .

F = 4.8 N, directed toward the origin

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Summary

  • A conservative force is one for which the work done is independent of path. Equivalently, a force is conservative if the work done over any closed path is zero.
  • A non-conservative force is one for which the work done depends on the path.
  • For a conservative force, the infinitesimal work is an exact differential. This implies conditions on the derivatives of the force’s components.
  • The component of a conservative force, in a particular direction, equals the negative of the derivative of the potential energy for that force, with respect to a displacement in that direction.

Conceptual questions

What is the physical meaning of a non-conservative force?

A force that takes energy away from the system that can’t be recovered if we were to reverse the action.

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A bottle rocket is shot straight up in the air with a speed 30 m/s . If the air resistance is ignored, the bottle would go up to a height of approximately 46 m . However, the rocket goes up to only 35 m before returning to the ground. What happened? Explain, giving only a qualitative response.

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An external force acts on a particle during a trip from one point to another and back to that same point. This particle is only effected by conservative forces. Does this particle’s kinetic energy and potential energy change as a result of this trip?

The change in kinetic energy is the net work. Since conservative forces are path independent, when you are back to the same point the kinetic and potential energies are exactly the same as the beginning. During the trip the total energy is conserved, but both the potential and kinetic energy change.

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Problems

A force F ( x ) = ( 3.0 / x ) N acts on a particle as it moves along the positive x -axis. (a) How much work does the force do on the particle as it moves from x = 2.0 m to x = 5.0 m? (b) Picking a convenient reference point of the potential energy to be zero at x = , find the potential energy for this force.

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A force F ( x ) = ( −5.0 x 2 + 7.0 x ) N acts on a particle. (a) How much work does the force do on the particle as it moves from x = 2.0 m to x = 5.0 m? (b) Picking a convenient reference point of the potential energy to be zero at x = , find the potential energy for this force.

a. −120 J ; b. 120 J

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Find the force corresponding to the potential energy U ( x ) = a / x + b / x 2 .

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The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by U ( x ) = a / x 12 b / x 6 where x is the distance between the atoms. (a) At what distance of seperation does the potential energy have a local minimum (not at x = ) ? (b) What is the force on an atom at this separation? (c) How does the force vary with the separation distance?

a. ( −2 a b ) 1 / 6 ; b. 0 ; c. x 6

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A particle of mass 2.0 kg moves under the influence of the force F ( x ) = ( 3 / x ) N . If its speed at x = 2.0 m is v = 6.0 m/s, what is its speed at x = 7.0 m?

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A particle of mass 2.0 kg moves under the influence of the force F ( x ) = ( −5 x 2 + 7 x ) N . If its speed at x = −4.0 m is v = 20.0 m/s, what is its speed at x = 4.0 m ?

14 m / s

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A crate on rollers is being pushed without frictional loss of energy across the floor of a freight car (see the following figure). The car is moving to the right with a constant speed v 0 . If the crate starts at rest relative to the freight car, then from the work-energy theorem, F d = m v 2 / 2 , where d , the distance the crate moves, and v , the speed of the crate, are both measured relative to the freight car. (a) To an observer at rest beside the tracks, what distance d is the crate pushed when it moves the distance d in the car? (b) What are the crate’s initial and final speeds v 0 and v as measured by the observer beside the tracks? (c) Show that F d = m ( v ) 2 / 2 m ( v 0 ) 2 / 2 and, consequently, that work is equal to the change in kinetic energy in both reference systems.

A drawing of a crate on rollers being pushed across the floor of a freight car. The crate has mass m,it is being pushed to the right with a force F, and the car has a velocity v sub zero to the right.
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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