Physics Homework Help?

Mass, m of the gymnast's feet is 100kg.

Work Done, W.D by the gymnast's feet to stop her is 5,000J, i.e. it takes the girl's feet 5,000J of energy to put her feet back to rest (Here we can say that initial velocity, u is greater than final velocity, v; thereby making v to be 0, i.e. v = 0.) So, our concern is initial velocity, u of the girl when she began to stop as this signifies that she's losing kinetic energy, K.E, and as a result of this the work done by her feet is considered to be negative because the cause of her work being done by her feet to put them back to rest is a retardating effect.

Using: change in kinetic energy, ΔK.E = - W.D, where

ΔK.E = ½ * mass, m * change in velocity - square, Δv²; then

½ * m * (v² - u²) = - 5,000

½ * 100 * (0 - u²) = - 5,000

50 * - u² = - 5,000

- 50u² = - 5,000

u² = 100

u = 10, wher u is measured in m/s, then

u = 10m/s ...Ans.

Since she is stopping she is losing KE. Therefore the work done takes energy out of the system and is negative. Change in KE = -work done

1/2*m*(vf^2-vi^2) = -work

Since vf = 0

-1/2*m*vi^2 = -work

1/2*m*vi^2 = work

Solve for vi

vi = sqrt(2*work / m)

Plug in numbers

vi = sqrt(2*5000 J / 100 kg) = 10 m/s