Solving a Mechanical Energy Problem Involving a Swing

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SUMMARY

The discussion focuses on calculating the mechanical energy dissipated by resistive forces in a swing problem involving a 42 kg child released from a height. The swing length is 2.3 m, and the initial angle with the vertical is 32 degrees. The gravitational acceleration is 9.8 m/s², and the child's speed at the lowest point is 2.34094 m/s. The correct calculation shows that the potential energy (PEg) at the release point is 501.66 J, and the kinetic energy (KE) at the lowest point is 115.08 J, resulting in a dissipated energy of 386.58 J.

PREREQUISITES
  • Understanding of gravitational potential energy (PEg) and kinetic energy (KE) equations
  • Basic trigonometry, specifically sine functions for angle calculations
  • Knowledge of energy conservation principles in mechanical systems
  • Ability to perform unit conversions and calculations in physics
NEXT STEPS
  • Review the principles of energy conservation in mechanical systems
  • Study trigonometric functions and their applications in physics problems
  • Learn about resistive forces and their impact on mechanical energy
  • Explore more complex swing dynamics and energy dissipation scenarios
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of energy dissipation in real-world scenarios.

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Homework Statement


A 42 kg child on a 2.3 m long swing is released from rest when the swing supports make an angle of 32 degrees with the vertical.
The acceleration of gravity is 9.8 meters per second squared. If the speed of the child at the lowest point is 2.34094 m/s, what is the mechanical energy dissipated by the various resistive forces (e.g. friction, etc.)? Answer in units of J

Homework Equations


PEg=mgh
KE=.5(m)(v^2)


3. The Attempt at a Solution
Attempted to solve using the lowest point of the swing as zero for potential energy. PEg at release point - KE at lowest point = dissipated energy.

PEg=mg(vertical component of swing length)
KE=.5(m)(v^2)

PEg=42(9.8)(2.3sin(32 degrees)
KE=.5(42)(2.34094^2)

PEg=501.66 J
KE=115.08 J

501.66-115.08=386.58 J =Answer

This however did not work..please help?
 
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You did not calculate the initial PE correctly. Draw a sketch and check your trig in determining how high above the low point the child starts.
 

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