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# Vertical Motion question.. watch

1. Hello, I have a vertical motion question which I dont seem to understand fully, though ive started it...

An acrobat of mass 'm' slides down a vertical rope of height 'h'. for the first three-quarters of her descent she grips the rope with her hands and legs so as to produce a frictional force equal to five-ninths of her weight. She then tightens her grip so that she comes to rest at the bottom of the rope. Find the frictional force she must produce in the last quarter. If the rope is 30m high, calculate
(a) Her greatest speed
(b) the time she takes to descend.

Could you at least help me with the first part...
(Finding this chapter really annoying)
Thanks a lot
From John
2. mg - 5/9*mg = m*a
4/9mg = m*a
a = 4/9g

a = 4/9g
u = 0
s = 3/4*30 = 22.5
v = ?

v2 = 2(as) = 196
v = 14ms-1(a)

a = ?
v = 0
u = 14
s = 1/4*30 = 7.5
196 + 15a = 0
a = -196/15-2

mg - f = m * -196/15
f = mg + m * 196/15 = 35*9.8/15m = 7/3mg (firctional force equal to 7/3 of her weight)

time of first trip = 14/(4/9g) = 45/14s
time of second trip = -14/(-196/15) = 15/14s
time of trip = 45/14 + 15/14 = 60/14 = 4.2857 = 4.29 (3 s.f.)
3. Thanks a lot, though I dont understand a lot how you got from

f = mg + m * 196/15 equal to that 35*9.8/15m = 7/3mg

Do u mind explaining..

Thanks
From John
4. f = mg + 196/15m
f = m(9.8+196/15)
f = 343/15m = 35*9.8/15m (simply divided 343 by 9.8)
f = 35/15 * mg = 7/3mg
6. Then just use s=(1/2)(u+v)(t) find time plot graph get total time . Remember the graph will be a v shape or alphabet upside down and find force using the same way you found your acceleration in the first step

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