My family got a new cat,
and during break I would often see it jump off the fence in the yard and into
the neighbor’s yard to socialize with their cats. The fence is pretty high, and
I was surprised that the cat never hurt itself jumping off of it. I thought
about the things that were reducing the force of the ground once the cat
reached it. The cat’s drag could be significant because it spread out and its coat
would get wider as it dropped. In addition to this, the length of the cat’s
legs also increased the distance traveled during the impact, which would
decrease the force.

Assuming that the cat’s total work done is:

^{}
Wnet = ΔKE + ΔPE – Wnc

Where Wnc is non-conservative work, and in this case is drag force. Plugging values in gives:

Wnet = ½ m(vf

It can be seen that the drag force decreases the magnitude of the net work done.

^{2}-vi^{2}) + mg (yf-yi) – Wnc = 1/2 (4 kg)((0 m/s)^{2}– (1 m/s)^{2}) + (4 kg)(9.8 m/s^{2})(0 m – 1.5 m) + 10 Nm = -2 Nm - 59 Nm + 10 Nm = -51 NmIt can be seen that the drag force decreases the magnitude of the net work done.

If the distance over which the work is done increases, the force will decrease. This can be seen in the equation:

F = Wnet/dcosθ

Assuming that the cat's legs are 0.25 m long, and the angle is 60 degrees, this gives an impact force of

F = -51 Nm/(0.25 m*cos60) = -408 N

Which is much smaller than what the impact force would be if the cat had not landed on its legs. This force is not difficult to withstand, and this example has not taken into account the absorption of impact from the cats muscles, which would further help decrease the impact force on the cat's bones and decrease the chance injury.

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