The Force Exerted by the Ground on a Parachuter

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SUMMARY

The discussion centers on calculating the force exerted by the ground on a parachutist who lands with an initial speed of 3.85 m/s and comes to rest over a distance of 0.700 m. The parachutist has a mass of 35.5 kg. The initial calculation using the equation Vf² = Vo² + 2ax yielded a deceleration of 10.5875 m/s², resulting in a force of 375.86 N. However, the correct approach requires adding the weight of the parachutist to this force, as the total force exerted by the ground must counteract both the deceleration and the gravitational force.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with kinematic equations, specifically Vf² = Vo² + 2ax
  • Knowledge of gravitational force calculations (weight = mg)
  • Basic algebra for solving equations
NEXT STEPS
  • Review the derivation and application of Newton's Second Law in various contexts
  • Study kinematic equations in detail to understand their applications in motion problems
  • Learn how to calculate net forces in scenarios involving multiple forces
  • Explore real-world applications of physics concepts in parachuting and landing dynamics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion and forces, particularly in the context of parachuting and landing scenarios.

proace360
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Homework Statement


A 35.5 kg parachutist lands moving straight downward with a speed of 3.85 m/s.
(a) If the parachutist comes to rest with constant acceleration over a distance of 0.700 m, what force does the ground exerts on her?

Variables: Vo, Vf, F, m, a, x
Known: Vo=3.85 (I think the problem means this as the initial)
m=35.5 kg
Vf=0 (don't quite understand the problem)
x=.7 m

Homework Equations


Fnet=m*anet
Possibly Vf^2=Vo^2+2ax

The Attempt at a Solution


I used Vf^2=Vo^2+2ax, and I got 10.5875 m/s^2, multiplied that by 35.5 kg, and I got 375.86 N. WebAssign says the answer is wrong.
 
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I figured it out. Apparently that is just the force required to stop the object--you must also factor in the weight.
 

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