Solving Compression Problem: Water Skier & Tow Rope

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The discussion revolves around calculating the change in length of a nylon tow rope used by a water skier under specific conditions. The skier experiences a resistive force of 190 N while being accelerated at 0.81 m/s². The net force acting on the skier is the difference between the force from the rope and the resistive force. The formula for elongation of the rope, Delta L = FL0 / (Y*A), is utilized to find the change in length. Understanding the net force is crucial for accurately determining the rope's elongation.
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A 57-kg water skier is being pulled by a nylon (Young's modulus 3.7 x 109 N/m2) tow rope that is attached to a boat. The unstretched length of the rope is 18 m and its cross-section area is 1.9 x 10-5 m2. As the skier moves, a resistive force (due to the water) of magnitude 190 N acts on her; this force is directed opposite to her motion. What is the change in length of the rope when the skier has an acceleration whose magnitude is 0.81 m/s2?


Ok here is what I was able to do so far I know that

F = Y(Delta L/L0)A

When I solve for Delta L I know that Delat L = FL0/ Y(A).

I just don't know how to factor in the acceleration and most importantly how to use the 190 N resistence force?

Please let me know and thanks a lot.

 
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The acceleration of the skier is the result of the net force acting. The net force is the combination of the force from the rope and the resistive force.
 
oh ic ic.

That explains it so much better.

Thank you so much.
 
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