Question

part a
Two rock climbers, Jim and Karen, use safety ropes of simi-lar length. Karen's rope is more elastic, called a dynamic rope by climbers. Jim has a static rope, not recommended for safety purposes in pro climbing. Karen falls freely about 2.0 m and then the rope stops her over a distance of 1.0 m. (Figure 1)
Part A
Estimate how large a force (assume constant) she will feel from the rope. Express the result in multiples of her weight.
Express your answer using two significant figures.
F =

Answer 1

Part B
In a similar fall, Jim's rope stretches by only 30 cm. How many times his weight will the rope pull on him?
Express your answer using two significant figures.
F =

Answer 2

@agent0smith

Answer 3

Karen falls freely about 2.0 m

use this to find the velocity before the rope starts to slow her... \[\Large v_f ^2= v_i ^2 + 2 a \Delta x\]
her inital velocity must be zero, a is gravity.

Answer 4

i got 2.. can you check that please

Answer 5

2 what...? Units are important.

Answer 6

2 is not correct.. but i don't even know what 2 is... velocity? force?

Answer 7

2N or is it -2N

Answer 8

sorry about that

Answer 9

First, her velocity would be 6.261 m/s. Then find her acceleration when the rope is slowing her, and use that to find force... which we can't, since we don't know her mass...??

Answer 10

Oh it's just in multiples of her weight

Answer 11

yes i got that answer, and then plugged it in the same formula to find a= 19.62m/s2

Answer 12

Correct.

Answer 13

then i took that a and divided it by 9.81m/s2 and got an answer of 2

Answer 14

2N

Answer 15

So it'd be 2 times her weight, then.

Answer 16

NOT 2 newtons.

Answer 17

oh then what would the units be?

Answer 18

2 times her weight. Or 2W i guess. It won't really have units, since the weight W has units of N, but 2W is just a ratio.

Answer 19

"Express your answer using two significant figures."

Answer 20

so it would be 2.0, i dont know why but something seems wrong

Answer 21

Why? 2 times her weight would be ~980N if she had a mass 50kg.

Answer 22

its wrong

Answer 23

The net force on her will be 2W.... but that's not the force from the rope

The force from the rope is T... which will be 3W

This might've been why the last question was wrong too... we forgot that it's only the NET force causing acceleration... the upward force has to also account for the weight, like the weight of the torso in the last problem.

Answer 24

In the last problem, the NET force decelerates the torso, but there's also the reaction force on the torso, which is just supporting the weight of the torso.

Answer 25

i got dont understand would you just add 1 to it then?

Answer 26

Solve the equation on the picture for T.

Answer 27

would i do 2W = T - W

T - 3W ?

Answer 28

T= 3W

Answer 29

Yes, T=3W

Answer 30

okay i got that, now i got 6.657 times his weight for part b i suspect that this is wrong too

Answer 31

we can complete this in another question spot if you want, if its getting to messy here

Answer 32

Okay. That one is basically the same as a anyway, with diff numbers. Open a new question