Finding rope extension with skier acceleration and water resistance

  • Thread starter Thread starter shaka23h
  • Start date Start date
  • Tags Tags
    Compression
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
2 replies · 5K views
shaka23h
Messages
38
Reaction score
0
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 resistance force?

Please let me know and thanks a lot.

 
Physics news on Phys.org

Discussion

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.