Physics

A 50.0 kg parachutist jumps out of an airplane at a height of 1.00 km. The parachute opens, and the jumper lands on the ground with a speed of 5.00 m/s. By what amount was the jumper's mechanical energy reduced due to air resistance during this jump?

so, you have to take the inital ME less the final ME, right? And since there is no motion initially the KE=0, and since the jumper is at its terminal velocity, there's no final PE, right? so then would it just be (PE_i - KE_f) or (50kg*9.81m/s^2*1000m)-(.5*50kg*(5^2 m/s))

I think it's incorrect because of the number I keep getting, but I don't know what I'm doing wrong.

(I'm home sick, I'll be here all day... so anytime you could answer this would be fine.)

I would assume the initial KE of the jumper is zero.

Your analysis is correct. Nearly all of the energy is dissipated in the air. Thankfully.

They may be wanting a percentage of energy dissipated, the wording is somewhat ambiguous.



I don't think it's a percentage, we haven't been working with percentages yet in this class.
So it'd be 490500-125?

also, it asks for 'What is the average force of air friction during the jumper's descent if the parachute opened immediately after jumping?' and I'm not sure what equation I could use for that situation.

Ok. Use

average force * distance traveled= energy dissipated.
solve for average force.

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asked by Becca
  1. pe=mgh
    (50)(9.81)(1000)=491000J
    KE=1/2mv^2= (.5)(50)(5)^2=625J
    Energy Lost= PE-KE=491000-625= 490375

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    posted by bo

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