Abig 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 itselfand hits the ground with outstretched legs decelerating to a stop over a distance of 12.0 cm (from rest on the branchat the top, to crouched at rest on the ground the total distancecovered is 2.00 m + 12.0 cm). assuming a constantdeceleration over the 12.0 cm, what force (in n) does the cat exerton the ground?

force exerted by the cat is 693 N

Explanation:

We know that total energy is the sum of work done by gravity and friction i.e.

solving for F

here -ve sign indicate work done in opposite direction

where, m is mass  of cat = 4.00 kg

h = 2.00 m,

Putting all value in above equation

F = 692.53 N = 693 N

Therefore force exerted by the cat is 693 N

d = Δv(t2-t1)

Explanation:

Speed is defined as the change of displacement with respect to time. It is expressed as shown;

Speed = change in displacement/change in time

Δv = d/Δt

d = Δv*Δt

d = ΔvΔt

Δt = t2-t1

d = Δv(t2-t1)

Δv is the change in rate of speed

Δt = change in time

The correct expression for the displacement of the car during this motion is d = Δv(t2-t1)