Cutnell, Physics,100 Helo I System Announcements (1 Unread) PRINTER VERSION BACK

Question

Cutnell, Physics,100 Helo I System Announcements (1 Unread) PRINTER VERSION BACK Chapter 10, Problem 55 A 53-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 13 m and its cross-section area is 3.0 x 10 s m2 As the skier moves, a resistive force (due to the water) of magnitude 110 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 0.58 m/s2 Number Units the tolerance is +/-3% Click if you would like to Show Work for this question: Open Show Work sHOW HINT LINK TO TEXT LINK TO TEXT By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor

Explanation / Answer

Fnet = ma

=> F – 110 = (53*0.58)

=> F = 140.74 N

We know,

E = stress/strain

=> E = (F/A)/(dL/L)

=> dL = FL/EA

= (140.74*13)/(3.7*10^9 * 3*10^-5)

= 0.01648m

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.