Some climbing ropes are designed to noticeably stretch under a load to lessen the forces in a fall. A weight is attached to 10 m length of climbing rope, which then stretches by 20 cm. Now, this single rope is replaced by a doubled rope--two pieces of rope next to each other. How much does the doubled rope stretch?

b. A 10 m length of climbing rope is supporting a climber, and stretches by 60 cm. When the climber is supported by a 20 m length of rope, by how much does the rope stretch?

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cjmejiab cjmejiab · Answer 1 · 2019-11-27T03:20:49+01:00

To solve this problem it is necessary to apply the concepts related to Hooke's law.

By definition the force through the theory formulated by Hooke is defined as

[tex]F = k \Delta X[/tex]

Where,

k = Spring constant

[tex]\Delta X =[/tex]Displacement

Part A)

[tex]F = k \Delta X[/tex]

[tex]k=\frac{F}{\Delta X}[/tex]

Therefore

[tex]\Delta X = \frac{F}{k} = 20m[/tex]

At this point our new spring constant would then be given by

[tex]k' = 2K[/tex]

[tex]\Delta X' = \frac{F}{2k} = \frac{20}{2}[/tex]

The expression that indicates how much has been displaced would be

[tex]\Delta X' = 10m[/tex]

Part B) If the length is doubled then the expression for the spring constant is

[tex]K'' = \frac{k}{2}[/tex]

Then the displacement would be

[tex]\Delta X' = 2\Delta X[/tex]

[tex]\Delta X' = 2*60[/tex]

[tex]\Delta X' = 120cm[/tex]