1
$\begingroup$

A vehicle moving with some velocity on a rough horizontal road finally comes to rest after its engine has been turned off. Intuitively, it seems a vehicle with greater mass would stop first because it would experience a greater friction force, but if we go by the work-energy theorem as follows, it's clear that the distance covered does not depend on the vehicle's mass at all:

$$K.E_{final} - K.E_{initial} = W_{friction}$$

$$KE_{final}=0\\ KE_{initial}=\frac12MV^2\\ W_{friction}=\mu Mgd\\ \text{Therefore, }d=\frac{v^2}{2\mu g}$$

Is my notion valid?

$\endgroup$
  • 1
    $\begingroup$ Learn some $LATEX$ to write mathematical equations at [here](meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) $\endgroup$ – RE60K Jul 10 '14 at 8:16
  • $\begingroup$ A heavier vehicle of the same dimensions will coast farther than a lighter vehicle. Both experience the same force from drag, but a heavier vehicle is less affected by it. There is no sliding friction because they use wheels and greater mass increases static friction with the tires on the road, which improves traction and distance travelled. $\endgroup$ – Jim Jul 10 '14 at 14:56
1
$\begingroup$

This is admittedly a late answer. Hopefully it will clear up some of the confusion.

If you ignore aerodynamic drag, ignore that the coefficient of rolling friction varies with load, and ignore a number of other factors such as friction between the axle and the bearings that support it, then yes, stopping time / stopping distance is independent of vehicle load. This is completely unrealistic reasoning.

Things get a bit more complex if you don't ignore those factors. Consider a pickup with or without a load of dirt. I'll start by assuming a constant coefficient of drag and by ignoring everything but aerodynamic drag. The only horizontal force acting on the vehicle is drag, and that is strictly proportional to velocity. Per these assumptions, a fully loaded pickup and empty pickup moving at the same speed suffer exactly the same drag force. Since acceleration is inversely proportional to mass, the loaded pickup takes longer to come to rest than the empty pickup.

Most people drive their pickups with the tailgate closed and the bed uncovered. This makes for a rather non-aerodynamic vehicle. Some people solve this by replacing the tailgate with an open mesh net, others solve this by adding a bed cover. I'll assume a typical pickup owner who hasn't done either. Filling the bed with dirt makes the vehicle more aerodynamic. This decrease in the coefficient of drag will make the loaded pickup take even longer to come to a stop than the simple analysis above.

The above ignores friction between the tires and the road. There will be a significant increase in contact area between tire and road after filling the pickup with dirt. This increased contact area increases the coefficient of rolling friction. This will act to make the loaded vehicle come to rest more quickly than the empty vehicle. Whether the loaded pickup has a longer or shorter stopping distance than the empty truck depends on which of aerodynamic drag or rolling friction predominate.

Anyone who has hauled a load of dirt with a pickup should know they need to reinflate the tires after taking on that load. Driving with a heavy load and tires squished flat to the ground is a good recipe for trouble. Reinflating the tires so that the contact area is more or less the same as it was prior to taking on the load reduces the coefficient of rolling friction. This makes drag and rolling friction work in concert with one another. The loaded truck takes a good deal more time or distance to come to a stop than does the empty truck.


Similar concerns sometimes come into play during the launch of a rocket. Rockets with a high thrust to weight ratio oftentimes need to coast for a while between the shutdown of one stage and the ignition of the next. When that's the case, the separation of the burnt-out stage doesn't occur when the stage shutdowns. That separation instead is delayed until shortly before the next stage is supposed to ignite. Keeping the burnt-out stage attached until the last minute reduces drag acceleration by a considerable margin.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Late, but equally helpful. Thanks, David. $\endgroup$ – Swami Sep 14 '14 at 11:48
0
$\begingroup$

Your intituition is totally different because ennumerous forces change the situation from the ideal situation predicted by work enrgy theorem, some of them are: Air Drag/Friction, Rotational Friction due to differences in size of tyres, different aerodynamic effects due to different body design.. etc.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You mean to say- it would certainly depend on mass as well besides other factors? $\endgroup$ – Swami Jul 10 '14 at 7:59
  • $\begingroup$ @Swami yes but intiution about greater force is wrong because even if the force is large, the mass will be compensatorily high to cancel effect of high force, and you'll result in equal acceleration; after all acceleration decides the motion $\endgroup$ – RE60K Jul 10 '14 at 8:08
  • $\begingroup$ That's still not very clear to me. $\endgroup$ – Swami Jul 10 '14 at 8:14
  • $\begingroup$ In a ideal world with only friction and gravity, distance will be same.Not will it be same in "our" real world. $\endgroup$ – RE60K Jul 10 '14 at 8:17
  • 1
    $\begingroup$ I'm not completely convinced. Suppose the only friction involved is between tire and road. Unless that force is a linear function of vehicle weight, the mass (and hence momentum) of the vehicle will matter. $\endgroup$ – Carl Witthoft Jul 10 '14 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.