Effects of velocity on weight (mass?)

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SUMMARY

The discussion centers on the relationship between velocity, kinetic energy, and the perceived weight of a driver during braking. When a vehicle traveling at 30 m/s experiences deceleration, the driver feels a force equivalent to more than 2g due to the acceleration of -22.5 m/s². Key formulas include kinetic energy (E = mv²/2) and momentum (momentum = mass * velocity). The confusion arises from the misuse of terms like "weight" and "mass," as mass remains constant regardless of velocity, while kinetic energy increases with speed, impacting the forces experienced during sudden stops.

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  • Understanding of Newton's laws of motion
  • Familiarity with basic physics concepts such as kinetic energy and momentum
  • Knowledge of acceleration and deceleration calculations
  • Ability to differentiate between mass and weight
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nuuskur
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A ballistics-esque question. Suppose we are driving at ##x \text{ m/s}## and we have to suddenly brake. For students in driving schools they are told something along the lines of "during the braking, the body experiences ##y \text{ kg}## of force" or "the mass of the body is much higher than at rest". Eventually they point out the dangers of driving without a seatbelt and that's all fine and dandy, but the technicalities are a bit odd to me. How do they come up with some specific numbers?

Suppose we drive at ##30 \text{ m/s} ##. If the driver's mass is ##80\text{ kg}##, then his kinetic energy would be computable by ##E = \frac{mv^2}{2}## yielding ##36 \text{ kJ} ##. How does one convert this to units of mass of an invisible body that is exerted on the driver during braking?

Alternatively we could also make use of
<br /> d = \frac{v^2 - v_0 ^2}{2a}<br />
where ##v## is the terminal velocity, ##v_0## is initial velocity, ##d## is distance and ##a## is acceleration. Suppose we have to come to a full stop in some ##20\text{ m}##, thus requiring an acceleration of ##-22.5 \text{ m/s}^2 ##. So the body would experience a bit more than ##2\text{ g}## of force (I'm not sure if this expression makes sense).

From the above it's not clear to me how one comes up with expressions of the form "at ##x \text{ km/h}## the driver weighs ##y## times more than at rest". Are there other lines of computations to be considered? What do they mean when they say "the body weighs ##y## units at velocity ##z## units" ?
 
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nuuskur said:
What do they mean when they say "the body weighs ##y## units at velocity ##z## units" ?

Ask "them". It has nothing to do with physics.
 
This is totally wrong. Mass does not change at all with velocity. Do you mean momentum? Momentum is calculated with the formula, ##momentum = mass * velocity##
They might also be talking about kinetic energy, which is very important when a car crashes. The kinetic energy is calculated with the formula, ##kineticEnergy = mass * velocity^2##.

PS. At speeds near the speed of light, some people misuse the word "mass" to describe something that increases at such huge speeds. That is not what you are talking about here when you talk about a car.
 
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One can argue that if the deceleration is 1g, the force exerted on you by the seatbelt is the same as if I used a crane to pick up the car by its tail, leaving you hanging from the seat held in by the straps. If the deceleration was 2g, the force is the same as if you were in that situation but with your twin brother lying on your back. 3g, your other twin brother joined him, etc.

Filtered through the inability of muggles to tell the difference between weight and mass, and adding the (not unreasonable) assumption that a high speed crash involves more brutal acceleration, I'd suspect that is what they're saying.
 
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