Answer :
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
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:
[tex]W=K_f -K_i[/tex] (1)
where
W is the work done
Ki is the initial kinetic energy
Kf is the final kinetic energy
In this problem, we have:
[tex]W=-5000 J[/tex] the net work done on the gymnast
[tex]m=100 kg[/tex] is the mass of the gymnast
[tex]v_f = 0[/tex] is the final velocity of the gymnast, so her final kinetic energy is also zero:
[tex]K_f = \frac{1}{2}mv_f^2 = 0[/tex]
Therefore, we can rewrite eq.(1) as
[tex]W=-\frac{1}{2}mv_i^2[/tex]
where [tex]v_i[/tex] is the initial velocity of the girl. By substituting the numbers and re-arranging the equation, we find:
[tex]v_i = \sqrt{-\frac{2W}{m}}=\sqrt{-\frac{2(-5000 J)}{100 kg}}=10 m/s[/tex]
