A 62-kg water skier is being pulled by a nylon (Young\'s modulus 3.7 x 10 9 N/m

Question

A 62-kg water skier is being pulled by a nylon (Young's modulus 3.7 x 109N/m2) tow rope that is attached to a boat. The unstretched length of the rope is 10 m and its cross-section area is 2.2 x 10-5m2. 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 0.49 m/s2?

HINT: Draw a free body diagram of the skier and use Newton's second law to find the tension in the rope.

Explanation / Answer

Force = ma + Resistance from water = 220.38 N

= Y A delta/L = 3.7 e9 x 2.2 e-5 x delta / 10

elongation = 27 mm

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