5. Misty May-Treanor (mass=66 kg) lands on the sand with a downward vertical vel

Question

5.     Misty May-Treanor (mass=66 kg) lands on the sand with a downward vertical velocity of 3 m/s. Her center of of gravity is displaced downward for .32 m from initial contact to complete stop.

a.     What is the change in kinetic energy from initial contact with the ground to complete stop?

b.     What is the change in potential energy of her center of gravity from initial contact with the ground to complete stop?

c.     How much work is done by the ground reaction force from initial contact to complete stop on Misty May?

d.     If the average force acting on Misty May from the ground during landing to complete stop is -247.5 N, what is the time it takes from landing to complete stop?

Explanation / Answer

a) change in kinetic energy = (1/2)*m*(vf^2 - vi^2)

= (1/2)*66*(0^2 - 3^2)

= -297 J

b) change in potential energy = -m*g*h

= -66*9.8*0.32

= -207 J

c) Workdone by the the ground = change in mechnical energy

= -297 - 207

= -504 J

d) he winever stops.

because, 247.5 N < m*g

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