Question

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 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?

Answer 1

force exerted by the cat is 693 N

Step-by-step explanation:

We know that total energy is the sum of work done by gravity and friction i.e.
`E = W_(gravtiy) + W_(friction)`

`0 = mg (h+s) + Fs`

solving for F

`F = -  (mg(h+s))/(s)`

here -ve sign indicate work done in opposite direction

where, m is mass of cat = 4.00 kg

h = 2.00 m,
`s = 12.00* 10^(-2) m`

Putting all value in above equation

`F = (4 * 9.87 *( 2 + 12* 10^(-2))/(12* 10^(-2))`

F = 692.53 N = 693 N

Therefore force exerted by the cat is 693 N