- #1
shaka23h
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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.
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.