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Example: The Hill

Probably the most classic example of how the WORK AND ENERGY law can be used is the case of a ball rolling down a frictionless hill, pictured schematically in Fig. 11.3.

  

Figure: Sketch of a ball rolling down a frictionless hill. In position 1, the ball is at rest. It is then given an infinitesimal nudge and starts to roll down the hill, passing position 2 on the way. At the bottom of the hill [position 3] it has its maximum speed  v₃ , which is then dissipated in rolling up the other side of the hill to position 4. Assuming that it stops on a slight slope at both ends, the ball will keep rolling back and forth forever.

Now, Galileo was fond of this example and could have given us a calculation of the final speed of the ball for the case of a straight-line path (i.e. the inclined plane); but he would have thrown up his hands at the picture shown in Fig. 11.3! Consider one spot on the downward slope, say position 2: the FBD of the ball is drawn in the expanded view, showing the two forces   $\mbox{\boldmath$\vec{N}$\unboldmath }$  and   $\mbox{\boldmath$\vec{W}$\unboldmath }$  acting on the mass  m  of the ball.^11.6 Now, the ball does not jump off the surface or burrow into it, so the motion is strictly tangential to the hill at every point.^11.7 Meanwhile, a frictionless surface cannot, by definition, exert any force parallel to the surface; this is why the normal force   $\mbox{\boldmath$\vec{N}$\unboldmath }$  is called a ``normal'' force - it is always normal [perpendicular] to the surface. So   $\mbox{\boldmath$\vec{N}$\unboldmath } \perp d\mbox{\boldmath$\vec{x}$\unboldmath }$  which means that   $\mbox{\boldmath$\vec{N}$\unboldmath } \cdot d\mbox{\boldmath$\vec{x}$\unboldmath } = 0$  and the normal force does no work ! This is an important general rule. Only the gravitational force   $\mbox{\boldmath$\vec{W}$\unboldmath }$  does any work on the mass  m, and since   $\mbox{\boldmath$\vec{W}$\unboldmath } = - m \, g \, \hat{y}$  is a constant downward vector [where we define the unit vector  $\hat{y}$  as ``up''], it is only the downward component of   $d\mbox{\boldmath$\vec{x}$\unboldmath }$  that produces any work at all. That is,   $\mbox{\boldmath$\vec{W}$\unboldmath } \cdot d\mbox{\boldmath$\vec{x}$\unboldmath } = - m \, g \, dy$, where  dy  is the component of  d $\mbox{\boldmath$\vec{x}$\unboldmath }$  directed upward.^11.8 That is, no matter what angle the hill makes with the vertical at any position, at that position the work done by gravity in raising the ball a differential height  dy  is given by   $dW = - m \, g \, dy$  [notice that gravity does negative work going uphill and positive work going downhill] and the net work done in raising the ball a total distance  $\Delta y$  is given by a rather easy integral:

$$\Delta W = - m \, g \int dy = - m \, g \, \Delta y$$

where  $\Delta y$  is the height that the ball is raised in the process. By our LAW, this must be equal to the change in the kinetic energy   $T \equiv {1\over2} m v^2$  so that

$${1\over2} m v^2 - {1\over2} m v_0^2 \; = \; - \, m \, g \, \Delta y .$$ (11.10)

This formula governs both uphill rolls, in which  $\Delta y$  is positive and the ball slows down, and downhill rolls in which  $\Delta y$  is negative and the ball speeds up. For the example shown in Fig. 11.3 we start at the top with   v₀ = v₁ = 0  and roll down to position 3, dropping the height by an amount  h  in the process, so that the maximum speed (at position 3) is given by

$${1\over2} m v_3^2 = m g h \qquad \hbox{\rm or} \qquad v_3 = \sqrt{2 g h} .$$

On the way up the other side the process exactly reverses itself [though the details may be completely different!] in that the altitude once again increases and the velocity drops back to zero.

The most pleasant consequence of this paradigm is that as long as the surface is truly frictionless, we never have to know any of the details about the descent to calculate the velocity at the bottom! The ball can drop straight down, it can roll up and down any number of little hills [as long as none of them are higher than its original position] or it can even roll through a tunnel or ``black box'' whose interior is hidden and unknown - and as long as I guarantee a frictionless surface you can be confident that it will come out the other end at the same speed as if it had just fallen the same vertical distance straight down. The direction of motion at the bottom will of course always be tangential to the surface.

For me it seems impossible to imagine the ball rolling up and down the hill without starting to think in terms of kinetic energy being stored up somehow and then automatically re-emerging from that storage as fresh kinetic energy. But I have already been indoctrinated into this way of thinking, so it is hard to know if this is really a compelling metaphor or just an extremely successful one. You be the judge. I will force myself to hold off talking about potential energy until I have covered the second prototypical example of the interplay between work and energy.