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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?