The federal government is considering raising the maximum weight for a heavy commercial vehicle from 80,000 pounds to 99,000 pounds. My question (Part 1) is: How much greater will the force be should a 99K truck hit the back of a stationary object vs the force of the 80K truck if both are traveling at 75 mph?

Part II: How much greater would the force be if both vehicles were traveling at 65 mph?

Part III: At what speed does the 99K truck need to be traveling to have an equal force to the 80K truck traveling at 75 and 65 mph.

Please be simple with your answers. I will be presenting this to the Senate Commerce Committee and they don't like complicated scenarios.

Many Thanks in advance

Tom Hodgson
Executive Director
Road Safe America

Tom, "force" is not necessarily the best measurement to look at here. The reason I say that is that for a given size truck, the force is directly proportional to the mass and the acceleration (which really means deceleration in English, but in Physics we call it acceleration even if it's slowing down). This is from Newton's 2nd law: F = ma. The reality is that upon impact, a heavier or faster-moving truck will take longer to stop, so the acceleration (and, therefore, force) may not be much (or any) greater than that of a slower-moving or lighter truck. My point is simply that force can be somewhat ambiguous and a little misleading in this case.

Bear in mind that the acceleration we're talking about here is that of a massive truck slowing from 65 or 75 mph to a complete stop (eventually). That acceleration is being facilitated by the sliding of tires on pavement, the crumpling of metal, the breaking of bones, etc. Clearly taking longer to stop is very bad, regardless of whether the force is higher or not.

That's why I would propose that you instead consider the energy of the collision. Ultimately, regardless of how slowly or quickly the truck comes to a stop, the kinetic energy of the truck prior to the collision must be absorbed by the brakes, crumple zones, air bags, trees, mailboxes, light posts, and (unfortunately) human bodies that happen to get in the way. The formula for kinetic energy is E_k=\frac 1 2 mv^2, where m is the mass (which is equivalent to weight in this case) and v is the velocity.

That means that for a given speed if one truck weighs twice as much as another (i.e. m is twice as big in the above formula), it has twice as much kinetic energy. That means twice as much energy must be absorbed by the truck and/or any stationary object it happens to slam into in the process of coming to a stop. On the other hand, if one truck travels twice as fast as another of the same weight (i.e. v is twice as big in the above formula), it's kinetic energy is four times as high!

Therefore, in going from 80k to 99k lbs, the energy of a truck will increase by a factor of 99/80 or around 24% (for a given speed). Likewise, for a given weight, increasing the speed from 65 mph to 75 mph will increase the energy by (\frac {75}{65})^2 or about 33%.

If you increase both mass and velocity, the energy increase is frightening large:

\Large \frac{E_{(99000lb , \; 75mph)}}{E_{(80000lb , \; 65mph)}}=\frac{99 \times 75^2}{80 \times 65^2} \approx 1.65

That's around a 65% increase in the kinetic energy of a truck! That's 65% more energy that must be absorbed by a stationary car on the side of the highway, along with its human occupants, in the case of a collision.

Also, if you want to do any of these calculations in "standard" units of energy, you'll probably want to do a few conversions (this is not necessary to see relative changes in energy like in the previous post, but will be helpful if you want to cite any examples where you calculate the actual energy dissipated):

(1) Express mass in kilograms (kg).

1 lb. weight = 0.454 kg mass

80,000 lb = (80,000 x 0.454) kg = 36300 kg
99,000 lb = (99,000 x 0.454) kg = 44900 kg

(2) Express speed in meters per second (m/s).

1 mph = 0.45 m/s

65 mph = 29.06 m/s
75 mph = 33.53 m/s

Using those standard units for mass and velocity will give you the energy in Joules (J). I know that's not a unit most people work with on a daily basis, but it IS the standard unit.

So in the case of a 99,000 lb. truck going 75 mph, we have

E=\frac 1 2 (44900)(33.53)^2\approx 50.5 \text { million Joules}

Finally, if you really want to drive you point home and make an impact (no puns intended), you should consider that a stick of dynamite releases about 2.1 million Joules of energy. That means the energy required to stop a 99,000 lb truck traveling at 75 mph is the equivalent of about 24 sticks of dynamite! (Whereas an 80,000 lb truck moving at 65 mph only generates as much energy as a mere 14.5 sticks).

jcaron2: First, thanks so much for such a detailed, yet understandable, answer. I asked for 'force' predicting I was using the wrong term. Thanks for helping me out. So, in very pedestrian language, adding 10% more weight is not as potentially damaging as adding 10% more speed. Right?

This I like, too. Do you teach or just think?

Yes, you're right that 10% more weight is not as bad as 10% more speed.

I'm not a teacher, but I'm always happy to help if the subject matter is one I happen to know well. :)

Tm: please do not use the comment feature of this web site to ask follow-up questions, but rather please use the answer box.

jcaron2's explanation is indeed first rate (as usual). But to answer your comment - yes, adding 10% to a vehicle's speed increases its kinetic energy by 21%, whereas adding 10% to the vehicle's mass while keeping its speed constant increases its energy by 10%.

By the way, to answer part 3 of your original qiuestion: if you increase the vehicle's weight from 80K to 99K (a 23.75% incease) then to have an equivalent amount of energy you need to decrease its speed to \sqrt{1-0.2375} = 90% of the original speed. So a 80K pound truck traveling at 65 MPH has the same energy as a 99K pound truck traveling at 58.4 MPH, and an 80K pound traveling at 75 MPH has the same energy as a 99K truck moving at 67.4 MPH. Hope this helps.