What is the maximum tension in the rope?

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SUMMARY

The maximum tension in the rope is calculated based on the dynamics of a 75 kg student swinging from a height after running at 5 m/s. The energy conservation principle is applied, where the kinetic energy at the start converts to potential energy at the peak height. The height (h) at which the student releases the rope is determined to be 1.276 meters. The tension in the rope just before release is derived from the forces acting on the student at that point.

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  • Understanding of energy conservation principles in physics
  • Familiarity with basic kinematic equations
  • Knowledge of forces and tension in ropes
  • Ability to perform calculations involving mass, velocity, and gravitational acceleration
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  • Study the concept of centripetal force in swinging motions
  • Learn how to apply the conservation of mechanical energy in different scenarios
  • Explore the dynamics of tension in ropes during pendulum-like movements
  • Investigate the effects of varying mass and speed on tension calculations
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Homework Statement


A 75 kg student runs at 5 m/s, grabs a rope, and swings out over a lake. He releases the rope when his velocity is zero. a) What is the angle when he releases the rope? b) What is the tension in the rope just before he releases it? c) What is the maximum tension in the rope?

Homework Equations


[tex]\Lambda[/tex]KE = [tex]\frac{1}{2}[/tex]mv[tex]_{f}[/tex][tex]^{2}[/tex] - [tex]\frac{1}{2}[/tex]mv[tex]_{o}[/tex][tex]^{2}[/tex] = mgh


The Attempt at a Solution



(75/2)*25 = 75 * 9.8 * h
h = 1.276

this is all I got.
 
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Was this the exact wording of from the question? I don't quite understand what it meant. First, you wrote when he releases the rope, his speed is 0; if so, there would not be energy transfer to the rope from the running speed of the student. Also ,have you provided all the info?
 

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