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A 76-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 19 m and its cross-section area is 1.6 x 10-5 m2. As the skier moves, a resistive force (due to the water) of magnitude 140 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 0.61 m/s2?

Respuesta :

Newton's second law and the definition of Young's modulus allows us to find the change in length in the chord is:

  • Length change is:  ΔL = 5.98cm

Given parameters

  • Mass of the water skier m = 76 kg
  • Young's modulus of chord Y = 3.7 10⁹ N / m²
  • Chord length Lo = 19 m.
  • The area of ​​the curda is a = 1.6 10-5 m²
  • Friction force with water fr = 140 N.
  • Acceleration a = 0.61 m / s²

To find

  • The change in length of the chord.

Newton's second law says that the net force on a body is directly proportional to its mass and acceleration.

           ∑ F = m a

where bold letters indicate vectors, F is force, m is mass, and acceleration.

A free-body diagram is a diagram of the forces without the details of the bodies, see attached.

          T-fr = m a

          T = ma + fr

Let's calculate.

          T = 76 0.61 + 140

          T = 186.36 N

Yung's modulus is defined by the relationship between stress and strain.

         

          [tex]Y = \frac{\frac{F}{A} }{\frac{\Delta L}{L_o} }[/tex]

           [tex]\Delta L = \frac{F}{A} \ \frac{L_o}{Y}[/tex]

Let's calculate.

 

         ΔL = [tex]\frac{186.36}{1.6 \ 10^{-5}} \ \frac{19}{3.7 \ 10^9}[/tex]  

         ΔL = 5.98 10⁻² m = 5.98 cm

In conclusion using Newton's second law and Young's modulus definition we can find the change in length in the chord is:

  • Length change is: ΔL = 5.98cm

Learn more about Young's Module here: brainly.com/question/13346614

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