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Mariulka [41]
3 years ago
10

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?
Physics
2 answers:
VikaD [51]3 years ago
6 0
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j 
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

yarga [219]3 years ago
6 0

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

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