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Force of a parachute on parachutist

  • Thread starter testme
  • Start date
  • #1
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Homework Statement


A parachutist of mass 65 kg jumps out of a plane deploys a chute.
When the chute opens, she slows down from 80km/h to 10km/h in 0.8s. What is the foce of the chute on her shoulders at that time?

m = 65 kg
v1 = 80km/h → 22 m/s
v2 = 10 km/h → 2.8 m/s
change in time = 0.8 s

Homework Equations



a = v/t
Fnet = ma

The Attempt at a Solution


We're assuming that up is positive.
I used to sig figs for all my values.

First we need to find the acceleration

a = v/t
a = (2.8 - 22)/0.8
a = -24 m/s^2

Let Fp = force of the parachute

Fnet = ma
Fg + Fp = ma
-640 + Fp = (65)(-24)
-640 + Fp = -1560
Fp = -1560 + 640
Fp = -920 N

I'm confused as to what I'm doing wrong. My teacher told us the answer should be about 2200 N and that would be positive while my answer is turning out completely off and negative.
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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473

Homework Statement


A parachutist of mass 65 kg jumps out of a plane deploys a chute.
When the chute opens, she slows down from 80km/h to 10km/h in 0.8s. What is the foce of the chute on her shoulders at that time?

m = 65 kg
v1 = 80km/h → 22 m/s
v2 = 10 km/h → 2.8 m/s
change in time = 0.8 s

Homework Equations



a = v/t
Fnet = ma

The Attempt at a Solution


We're assuming that up is positive.
I used to sig figs for all my values.

First we need to find the acceleration

a = v/t
a = (2.8 - 22)/0.8
a = -24 m/s^2
by using positive values for v, you have assumed down is positive and up is negative
Let Fp = force of the parachute

Fnet = ma
Fg + Fp = ma
-640 + Fp = (65)(-24)
-640 + Fp = -1560
Fp = -1560 + 640
Fp = -920 N
since the acceleration is up, the net force must be up; thus, Fp > Fg. Correct your plus and minus signs.
I'm confused as to what I'm doing wrong. My teacher told us the answer should be about 2200 N and that would be positive while my answer is turning out completely off and negative.
Net force is always in the direction of the acceleration.
 
  • #3
68
0
Okay, so then I changed the velocity values to negatives. That in turn made the acceleration a positive and then following through with

Fnet = ma
Fg + Fp = ma
-640 + Fp = (65)(24)
Fp = 2200 N

Thanks :)
 

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