Question

a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem to find the girl's initial velocity when she began to stop?

Answer 1

W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

Answer 2

10 m/s

Step-by-step explanation:

The work-kinetic energy theorem states that the net work done on an object is equal to the change in the kinetic energy of the object.

In formula:

`W=K_f -K_i` (1)

where

W is the work done

Ki is the initial kinetic energy

Kf is the final kinetic energy

In this problem, we have:

`W=-5000 J` the net work done on the gymnast

`m=100 kg` is the mass of the gymnast

`v_f = 0` is the final velocity of the gymnast, so her final kinetic energy is also zero:

`K_f = (1)/(2)mv_f^2 = 0`

Therefore, we can rewrite eq.(1) as

`W=-(1)/(2)mv_i^2`

where
`v_i` is the initial velocity of the girl. By substituting the numbers and re-arranging the equation, we find:

`v_i = \sqrt{-(2W)/(m)}=\sqrt{-(2(-5000 J))/(100 kg)}=10 m/s`