Why a heavier skier/whatever is faster, the answer.

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Heavier objects, such as skiers or cyclists, fall faster due to the impact of air resistance, which affects acceleration. While gravity accelerates all objects equally, the influence of drag force diminishes with increased mass. The acceleration formula shows that heavier skiers experience less relative air resistance, allowing them to maintain higher speeds. Friction also plays a role, but its effect is independent of mass. Consequently, a heavier skier will ultimately be faster on a slope.
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This doesn't directly belong in homework help, but I was trying to answer this question and realized it's rarely well answered.

The question is, why does a heavier ball, heavier skier, cyclist, fall faster? Many people seem to believe they will fall at the same speed, as acceleration due to gravity is the same for all of them, however common sense suggests this is wrong. A metal feather WILL fall faster than a real one. Why is this?

The simple answer is air resistance, however how we get to this takes a bit longer. We may also ask the question, doesn't friction also play a part?

With the example of a skier:

The acceleration of the skier at a given time is given by:

a = \frac{\sum F}{m}

\sum F = F_wsin \phi + F_d + F_f

Where phi is the angle of the slope, Fd is the drag force (air resistance), and Ff is the frictional force.

F_w = mg

F_d = -\rho A C_d v^2

http://en.wikipedia.org/wiki/Drag_equation"

F_f = -\mu F_wcos\phi

From this we can get:

a = \frac{\sum F}{m} = \frac{F_w + F_f + F_d}{m} = \frac{mg\sin \phi - F_w \mu \cos \phi - \rho AC_dv^2}{m}<br />

So

a = g \sin \phi - \mu g \cos \phi - \frac{\rho A C_dv^2}{m}

From this we can see that acceleration due to gravity and frictional force are independent of the mass, but the effect of air resistance on acceleration decreases with a larger mass!

Therefore, a heavier skier will be faster!


This might not be the right forum for this, feel free to move it.
If anyone thinks I'm wrong, or can explain it better, please post!
 
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a = g \sin \phi - \mu g \cos \phi - \frac{\rho A C_dv^2}{m}

but \rho = m / V
where V is the volume of the skier.

So,

a = g \sin \phi - \mu g \cos \phi - \frac{ A C_dv^2}{V}

which is independent of the mass of the skier.
 
rho is the density of the fluid, in this case the air, not the skier.

It's a constant.
 
re1s said:
rho is the density of the fluid, in this case the air, not the skier.

It's a constant.

Yeah, I realized that about 1 minute before you posted. Mea culpa.
 
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