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

Answer:

10 m/s

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:

$K_f = \frac{1}{2}mv_f^2 = 0$

Therefore, we can rewrite eq.(1) as

$W=-\frac{1}{2}mv_i^2$

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

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