A big clumsy cat having a mass of 4.00 kg falls from a tree branch for a distance of 2.00 m. In the air it rights itself and hits the ground with outstretched legs decelerating to a stop over a distance of 11.0 cm (from rest on the branch at the top, to crouched at rest on the ground the total distance covered is 2.00 m + 11.0 cm). Assuming a constant deceleration over the 11.0 cm, what force (in N) does the cat exert on the ground? (Hints: a kinematic equation used twice will get you most of the way there, and don’t forget about the cat’s weight.)

F = 752.68 N (force required to stop)

Explanation:

Let's first find the total speed the cat was travelling at when it hit the ground.

This can be calculated by the following motion equation:

Here u = initial velocity = 0

a = acceleration due to gravity = 9.81 m/s

s = Distance covered from branch to ground = 2 m

Solving for v , we get 6.2641 m/s.

Let's solve the same motion equation again to find the deceleration required:

v^2 - u^2 = 2*a*s

where u = initial velocity = 6.2641 m/s

v = final velocity = 0 m/s

s = 0.11 m or 11 cm

Solving for a, we get - 178.36 m/s^2

Finally we can plug this acceleration into Newton's force equation : F = m*a

The force needs to incorporate the weight of the cat as well, so we can divide this into:

F1 = weight of cat = 4 * 9.81 = 39.24

F2 = average force required to stop

Newton's equation now becomes:

F1 - F2 = mass * acceleration

39.24 - F2 = 4 * (-178.36)

F2 = 752.68 N (force required to stop)

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation: