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  4. Potential energy and conservation
  5. Conservation of energy

8.3 Conservation of energy  (Page 4/9)

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Check Your Understanding You probably recall that, neglecting air resistance, if you throw a projectile straight up, the time it takes to reach its maximum height equals the time it takes to fall from the maximum height back to the starting height. Suppose you cannot neglect air resistance, as in [link] . Is the time the projectile takes to go up (a) greater than, (b) less than, or (c) equal to the time it takes to come back down? Explain.

b. At any given height, the gravitational potential energy is the same going up or down, but the kinetic energy is less going down than going up, since air resistance is dissipative and does negative work. Therefore, at any height, the speed going down is less than the speed going up, so it must take a longer time to go down than to go up.

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In these examples, we were able to use conservation of energy to calculate the speed of a particle just at particular points in its motion. But the method of analyzing particle motion, starting from energy conservation, is more powerful than that. More advanced treatments of the theory of mechanics allow you to calculate the full time dependence of a particle’s motion, for a given potential energy. In fact, it is often the case that a better model for particle motion is provided by the form of its kinetic and potential energies, rather than an equation for force acting on it. (This is especially true for the quantum mechanical description of particles like electrons or atoms.)

We can illustrate some of the simplest features of this energy-based approach by considering a particle in one-dimensional motion, with potential energy U ( x ) and no non-conservative interactions present. [link] and the definition of velocity require

K = 1 2 m v 2 = E − U ( x ) v = d x d t = 2 ( E − U ( x ) ) m .

Separate the variables x and t and integrate, from an initial time t = 0 to an arbitrary time, to get

t = ∫ 0 t d t = ∫ x 0 x d t 2 [ E − U ( x ) ] / m .

If you can do the integral in [link] , then you can solve for x as a function of t .

Constant acceleration

Use the potential energy U ( x ) = − E ( x / x 0 ) , for E > 0 , in [link] to find the position x of a particle as a function of time t .

Strategy

Since we know how the potential energy changes as a function of x , we can substitute for U ( x ) in [link] , integrate, and then solve for x . This results in an expression of x as a function of time with constants of energy E , mass m , and the initial position x 0 .

Solution

Following the first two suggested steps in the above strategy,

t = ∫ x 0 x d x ( 2 E / m x 0 ) ( x 0 − x ) = 1 ( 2 E / m x 0 ) | −2 ( x 0 − x ) | x 0 x = − 2 ( x 0 − x ) ( 2 E / m x 0 ) .

Solving for the position, we obtain x ( t ) = x 0 − 1 2 ( E / m x 0 ) t 2 .

Significance

The position as a function of time, for this potential, represents one-dimensional motion with constant acceleration, a = ( E / m x 0 ) , starting at rest from position x 0 . This is not so surprising, since this is a potential energy for a constant force, F = − d U / d x = E / x 0 , and a = F / m .

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Check Your Understanding What potential energy U ( x ) can you substitute in [link] that will result in motion with constant velocity of 2 m/s for a particle of mass 1 kg and mechanical energy 1 J?

constant U ( x ) = −1 J

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We will look at another more physically appropriate example of the use of [link] after we have explored some further implications that can be drawn from the functional form of a particle’s potential energy.
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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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