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At a circus, an acrobat is being lifted straight upwards by a long wire. Though

ID: 1544402 • Letter: A

Question

At a circus, an acrobat is being lifted straight upwards by a long wire. Though the acrobat is currently slowing down as she nears the top of the circus tent, her custume includes layers designed to catch the air as she rises, so air resistance should not be ignored.

a) Crate a free body diagram for the acrobat

b) Starting from Fnet=ma, write an equation that replaces "Fnet" with the actual sum of forces acting on the acrobat. Symbols onlys no numbers

c) The acrobat's mass is 58kg, she's slowing down at a rate of 2.3 m/s2 , and the tension in the wire is 460 N. Starting from your equation in b, determine the magnitude of air resistance force acting on her.

d) At this instant, she's below 18m the top of the tent and her speed is 8.3 m/s. When she stops, how far below the top of the tent will she be?

Explanation / Answer

a)

b)

Fnet = ma

mg + Fr - T = ma               since acrobat is slowing down

c)

m = 58 kg

T = 460 N

a= 2.3

using the equation

mg + Fr - T = ma

58 x 9.8 + Fr - 460 = 58 (2.3)

Fr = 25 N

d)

Vo = initial speed = 8.3 m/s

Vf = final speed = 0

a = acceleration = - 2.3

d = distance travelled before stopping

using the equation

Vf2 = Vo2 + 2 a d

02 = 8.32 + 2 (-2.3) d

d = 14.98 m

distance from top = h = H - d = 18 - 14.98 = 3.02 m

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