#1

I'm in AP Physics B and am having a little trouble with forces.

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?

Now I understand that the force is going to be equal to the force with which she is pulled down, but how would I be able to find this force? I've tried 9.8 m/s^2 as her acceleration, but I don't understand if the that speed is the final speed or the speed before the .7 meters... basically I have no clue what is happening except that her acceleration isn't 0 or 9.8.

Any help would be greatly appreciated.

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?

Now I understand that the force is going to be equal to the force with which she is pulled down, but how would I be able to find this force? I've tried 9.8 m/s^2 as her acceleration, but I don't understand if the that speed is the final speed or the speed before the .7 meters... basically I have no clue what is happening except that her acceleration isn't 0 or 9.8.

Any help would be greatly appreciated.

#2

=56

#3

I know this is the Pit, but I think that it isn't too much to ask for serious answers

#4

If the acceleration isn't 0 and it's just gravity acting (IE: no other acceleration is mentioned in the y direction), the acceleration is 9.8 m/s^2.

#5

The acceleration is a deceleration, she's slowing down from 3.85 m/s to 0 in the space of 0.700m, so you can work out her speed as she hits the floor, then the force upward is equal to the force exerted by her landing.

Not actually sure that makes any sense

Not actually sure that makes any sense

#6

her final velocity is 3.85 m/s, and her initial velocity was 0. you know d, so vf^2 - vi^2=2ad (final velocity squared minus initial velocity squared = 2(acceleration)(distance). This is one of the four kinematic eqns. solve for a, then you can use that to find her force exerted (a x 35.5)

#7

how was her initial 0? I'm sure she didn't parachute for only.7 meters

#8

The acceleration is a deceleration, she's slowing down from 3.85 m/s to 0 in the space of 0.700m, so you can work out her speed as she hits the floor, then the force upward is equal to the force exerted by her landing.

Not actually sure that makes any sense

Thats what I did. 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. Still the wrong answer though.

*Last edited by orcs4life at Sep 21, 2008,*

#9

Since she's got constant deceleration, you can half her speed and work out how long it would take to travel .7m at that speed to get the time. Then divide her initial speed by that number to get her deceleration. Then multiply by her mass to get the force.

#10

I might do it as an impulse problem

Impulse=FnetT=m(v2-v1)=p2-p1

you know the change in velocity and the mass so that is equal to the time multiplied by the net force so just divide by time to get Fnet isolated and solve, you can use kinematics formulas to find the acceleration and time and just use that with or without the above formula.

Impulse=FnetT=m(v2-v1)=p2-p1

you know the change in velocity and the mass so that is equal to the time multiplied by the net force so just divide by time to get Fnet isolated and solve, you can use kinematics formulas to find the acceleration and time and just use that with or without the above formula.

#11

Since she's got constant deceleration, you can half her speed and work out how long it would take to travel .7m at that speed to get the time. Then divide her initial speed by that number to get her deceleration. Then multiply by her mass to get the force.

I tried that and you get the same answer as the above mentioned equation.

#12

I might do it as an impulse problem

Impulse=F(net)T=m(v2-v1)=p2-p1

you know the change in velocity and the mass so that is equal to the time multiplied by the net force so just divide by time to get Fnet isolated and solve, you can use kinematics formulas to find the acceleration and time and just use that with or without the above formula.

We've never learned impulse problems.

#13

We've never learned impulse problems.

Just completely skip the formula and use kinematics to find acceleration and multiply that by mass to get a net force, thats probably faster anyways.

#14

I might do it as an impulse problem

Impulse=FnetT=m(v2-v1)=p2-p1

you know the change in velocity and the mass so that is equal to the time multiplied by the net force so just divide by time to get Fnet isolated and solve, you can use kinematics formulas to find the acceleration and time and just use that with or without the above formula.

....I ****ing love your love for Clone High. Best show EVAR.

#15

^

slowing down from 3.85m/s to 0 over 0.7m with a constant acceleration requires it to be 10.5875m/s^2(down) so I think the answer your book gives might be wrong or the question is asking something else.

slowing down from 3.85m/s to 0 over 0.7m with a constant acceleration requires it to be 10.5875m/s^2(down) so I think the answer your book gives might be wrong or the question is asking something else.

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