What is the initialmechanical energy (sum of the kinetic and potential energies)

Question

What is the initialmechanical energy (sum of the kinetic and potential energies)? Whatis the final mechanical energy (when the particle stops somewhereon the horizontal track)? So, how much energy is transferred tothermal energy along the horizontal track? What are the magnitudesof the normal force and the kinetic frictional force? How is theenergy lost to thermal energy related to the frictional force andthe total distance of sliding? To slide that total distance, wheredoes the particle end up?

Explanation / Answer

The particle starts with a height of h= L/2 = 40 cm /2 = 20 cm. The particle starts from rest, so the initial kinetic energy is 0,and the initial potential energy is mgh = mgL/2 (since h =L/2). When the particle stops somewhere on the horizontal track, it'smechanical energy is 0 (it's stopped, so kinetic is 0, and it's onthe horizontal part so h = 0 giving 0 for potential energy aswell). So initial mechanical energy is non-zero, while final mechanicalenergy is zero. This means that all of the initial mechanicalenergy is transferred to thermal energy along the horizontal track.So amount of mechanical energy transferred to thermal energy is mgh= mgL/2 (all of it is transferred). The normal force keeps the particle from falling through the track,so in this case the normal force balances the force of gravity.FN = Fg = mg where FN is thenormal force, Fg is the force of gravity, m is the massof the particle, and g is acceleration due to gravity. Magnitude ofthe normal force is thus mg. Magnitude of kinetic frictional force: Fk = kFN = 0.29 FN = 0.29mg The energy lost to thermal energy comes from the work done on theparticle by the frictional force. Work is force times distance. Sowork done by friction will be the kinetic frictional force timesthe distance the particle slides. Wk = Fkd = 0.29mgd but we know that ALL of the mechanical energy is eventuallyconverted to thermal energy, so the work done by friction is equalto the initial mechanical energy of the system. Thus mgL/2 = 0.29 mgd cancel out m and g giving L/2 = 0.29d so d = L / (2*0.29) and remember L = 40 cm. plug 40 in for L and solvethis equation for d. (You'll get d = 68 point something--you can get the actual answeryourself) One more thing to note here: d is greater than the length of thehorizontal part. So the particle hasto start sliding back to theleft before it actually stops (because it has to slide about 68 cmbefore stopping). So it will slide 40 cm to the right, and then itwill slide 68 - 40 = about 28 cm back to the left (starting fromthe right side). That means it will end up 40-28 cm from the leftside (about 12 cm from the left side). *Note that I didn't use any decimal places for d, but it actuallyhas some that you should use. keep these decimal places in all ofyour calculations.

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