Kinetic and Potential Energy Advanced

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Contents

Introduction

In Work Introduction the basic concept of work and energy were defined, in Kinetic and Potential Energy, the concepts were expanded to discuss the various forms of energy in motion. Now, we will take these concepts further emphasizing the conservation of energy in all systems and expanding yet again the various forms of energy to include the potential energy found in spring and planetary bodies.

The Potential Energy in Springs (Elastic Potential Energy)

(Refer to Springs for the basics of forces caused by springs.) In the previous section we discussed how an object raised in the air contained potential energy because once it is let go, gravity performs work on it and the object accelerates. Consider the same object, this time, pushed against a spring. Once let go, the spring pushes the object, accelerating it in the direction of the spring expansion. For the same reasons then, an object pushed against a spring contains potential energy. How much energy? In the case of our W=Fd formula, or PE=mgh formula, the force felt is constant (F is constant in W=Fd and mg is constant in PE=mgh). But what about the spring? From previous discussions of springs, we know the force a spring exerts is F=-kx, where k is the spring constant and x is the distance is compressed from equilibrium. Therefore, we have a problem, we can't just use W=Fd to get the total potential energy that would be exerted from the spring because F keeps changing! The long answer to this is that we must integrate the changing force across the displacement (a straight line) however, in short, integration is beyond the scope of the MCAT and therefore not necessary to remember unless you would prefer to understand why the formula's are the way they are. The total potential energy stored in a spring then is

PE_{spring} = \frac{1}{2}kx^{2}

As with any other types of energy, we can use the conservation of total energy to determine how a system changes. For instance, an object pushed against a spring, when released, will convert this potential energy into kinetic energy (just as a falling object converted its gravitational potential energy into kinetic energy).


1. A cart with mass 1.50 kg is traveling 2.00 m/s on a frictionless track when it contacts an ideal spring with constant k = 3750 N/m. How far does the spring compress from equilibrium when the cart comes to a momentary stop?

2 cm
4 cm
2 m
4 m
→ This problem is a typical example of one that can be solved using KE, PE, and work. In this case, we can consider the total energy of the system (the moving cart and spring) before and after the collision.
→ Before, there is only one energy, the KE of the moving cart.
→ \begin{array}{lcl} KE_{initial} &=& \frac{1}{2}m_{cart}v_{initial}^{2} \\ &=& \frac{1}{2}\left(1.5 kg \right)\left( 2 m/s \right)^{2} \\ \end{array}
→ After the collision, the moving cart will come to a complete stop and the spring will become compressed converting all the kinetic energy to potential elastic energy.
→ Therefore, for the spring we have,
→ PE_{final} = \frac{1}{2}k_{spring}x^{2}
→ From the conservation of energy we have,
→ KE_{initial} + PE_{initial} = KE_{final} + PE_{final}
→ or,
→ \begin{array}{lcl} KE_{cart} &=& PE_{spring} \\ \frac{1}{2}m_{cart}v_{initial}^{2} &=& \frac{1}{2}k_{spring}x^{2}
→ solving for x then we get, that x2=0.0016, or x = 0.02m (20 cm).

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Gravitational Potential Energy (Grand Scale)

In the previous sections, when the concept of gravitational potential energy was introduced with the formula, PE = mgh, this was not entirely accurate. PE = mgh is only applicable when gravity is constant (and usually 9.8m/s2), as is the case with most problems involving projectiles, falling masses, incline planes, and objects being dragged on surfaces. In short, mgh only works when we are working with short distances/heights where gravity does not change. But we know that the force of gravity DOES change over large distances. When we deal with problems involving planets, stars, and rocket ships moving across space where the forces between them keep changing we must resort to a more general formula.

In Newton mechanics we learned that the force felt between two objects in space is

F = G \frac{m_1 m_2}{r^2}

Again, using an integration to determine the area of a force vs distance plot, we can derive how much potential energy is in this system (considering the fact that the strength of the force keeps changing with distance)

U = -G \frac{m_1 m_2}{R}


When gravitational potential energy is involved, keep the following distinction in mind,


Path Dependent, Independent, and Conservative Forces

Now that our understanding of the various forms of energy is complete, a few concepts remain to tie everything together. The first is that of path dependent and independent forces. To explain these concepts, let us consider two examples. In the first example we will raise an object up in the air doing work against gravity, in the second, we will drag a box along the ground doing work against friction.

Lets consider two different ways to raise an object in the air. We could lift the box straight up over our heads or we could pick up the object, walk over to and up a staircase until the object reached the same height. From our PE=mgh formula, h is the same, and so we have done the same amount of work in overcoming gravity. The path we took to do it (either straight up or against gravity) really didn't matter as both in the end had the same change in height. This is considered a path-independent force because the only significant information we need to determine the energy requirements is the initial and final position (and not the path in between them).

Now let us consider the second example. To move a box along the ground from point A to point B, we could drag it along the ground in a straight line, or perhaps we could drag it along a semicircular path. With either path we take, friction will be resisting us. But friction acts along the entire path and so a longer path (the semicircle) will experience more friction! This is known as a path-dependent force because the total amount of work/energy that is used in the end to overcome this force depends not just on the starting and ending position (displacement) but on the complete path traveled.

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