Newtonian Mechanics :)

Classical Mechanics Level 5

A person of mass 80 kg jumps from a height of 1 meter and foolishly forgets to buckle his knees as he lands. His body decelerates over a distance of only one cm. Calculate the total force on his legs during deceleration.

Details

  • g g = 10 m/s 2 ^2


The answer is 80800.

Fares Salem by Fares Salem , 22, Egypt

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2 solutions

Divyanshu Vadehra
Oct 10, 2014

i think that the answer should be 80800 and not 8080.This is how- decelaration=v^2-u^2/2S(third equation of motion) putting v=0m/s, u=underoot 2gh m/s=underoot 20m/s, s=0.01m, we get decelaration=1000m/s^2

Thus, pseudo force(acting downwards)=ma=80*1000=80,000N Total force=ma+mg=80,000+800=80800 and not 8080.Quite confused!!

7 Helpful 0 Interesting 0 Brilliant 0 Confused

I'm sorry guys for the delay in answering your nice questions; Here's the solution my friends: The person has mechanical energy E1 = mg(h+s) just before he lands the work done by him during deceleration is E2 = (f)(s) , (f): the total force on his leg. As E1 = E2 Therefore : f = [(m)(g)(h)] ÷ (s) + [(m)(g)] = [(80)(1)] ÷ (0.01) + [80] g = 8080g N

Fares Salem - 6 years, 7 months ago

i am satisfied by your answer

Yogesh Ghadge - 6 years, 7 months ago

perfectly correct.........

amey chaturvedi - 6 years, 7 months ago

hey! i believe you are correct

Ashu Dablo - 6 years, 7 months ago
Fares Salem
Oct 29, 2014

I'm sorry guys for the delay in answering your nice questions; Here's the solution my friends:

The person has mechanical energy E1 = mg(h+s) just before he lands the work done by him during deceleration is E2 = (f)(s) , (f): the total force on his leg.

As E1 = E2

Therefore :

f = [(m)(g)(h)] ÷ (s) + [(m)(g)]

= [(80)(1)] ÷ (0.01) + [80] g

= 8080g N

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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