Editing Work (physics) (section)

===Coasting down an inclined surface (gravity racing)===
Consider the case of a vehicle that starts at rest and coasts down an inclined surface (such as mountain road), the work–energy principle helps compute the minimum distance that the vehicle travels to reach a velocity {{math|''V''}}, of say 60&nbsp;mph (88 fps).  Rolling resistance and air drag will slow the vehicle down so the actual distance will be greater than if these forces are neglected.

Let the trajectory of the vehicle following the road be {{math|'''X'''(''t'')}} which is a curve in three-dimensional space.  The force acting on the vehicle that pushes it down the road is the constant force of gravity {{math|1='''F''' = (0, 0, ''w'')}}, while the force of the road on the vehicle is the constraint force {{math|'''R'''}}.  Newton's second law yields,
<math display="block"> \mathbf{F} + \mathbf{R} = m \ddot{\mathbf{X}}. </math>
The [[scalar product]] of this equation with the velocity, {{math|1='''V''' = (''v''<sub>x</sub>, ''v''<sub>y</sub>, ''v''<sub>z</sub>)}}, yields
<math display="block"> w v_z = m\dot{V}V,</math>
where {{math|''V''}} is the magnitude of {{math|'''V'''}}.  The constraint forces between the vehicle and the road cancel from this equation because {{math|1='''R''' ⋅ '''V''' = 0}}, which means they do no work.
Integrate both sides to obtain
<math display="block"> \int_{t_1}^{t_2} w v_z dt = \frac{m}{2} V^2(t_2) - \frac{m}{2} V^2 (t_1). </math>
The weight force ''w'' is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,
<math display="block"> w \Delta z = \frac{m}{2}V^2. </math>
Recall that V(''t''<sub>1</sub>)=0. Notice that this result does not depend on the shape of the road followed by the vehicle.

In order to determine the distance along the road assume the downgrade is 6%, which is a steep road.  This means the altitude decreases 6 feet for every 100 feet traveled—for angles this small the sin and tan functions are approximately equal.  Therefore, the distance {{mvar|s}} in feet down a 6% grade to reach the velocity {{mvar|V}} is at least
<math display="block"> s = \frac{\Delta z}{0.06}= 8.3\frac{V^2}{g},\quad\text{or}\quad s=8.3\frac{88^2}{32.2}\approx 2000\mathrm{ft}.</math>
This formula uses the fact that the weight of the vehicle is {{math|1=''w'' = ''mg''}}.