Radial Track, Find Loss of TME

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A 500g block slides down a radial track with a height of 1.0 m and a radius of 1.0 m, reaching a velocity of 4 m/s at the bottom. The total mechanical energy (TME) loss is calculated using the equation TMEf - TMEo, factoring in kinetic and potential energy changes. The initial calculations incorrectly used Newtons for energy and did not convert mass to kilograms, leading to confusion. The corrected equation shows that the loss of TME is 900 J, confirming the energy dissipation due to friction. The discussion emphasizes the importance of proper unit conversion and energy measurement in solving such problems.
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looks like a fun problem, just got to understand it:

A 500g block slides down a radial track with height of 1.0 m, it's radius is 1.0 m, with velocity of 4 m/s at the end of the track. what is the loss of the TME?

Disipation of energy problem:
TMEf - TMEo
F * D = K.E.f + P.E.f - (K.E.o + P.E. o)
friction * cos(180) * D = (1/2 mv^2f + mghfo) - (1/2mv^2o + mgho)
f(-1)(1) = (1/2)(500)(4)^2 + 0 - (500)(9.8)(1)

Can someone please confirm this?
is the loss of TME = 900 N = Friction ??
 
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Energy isn't measured in Newtons and you haven't converted your mass to kilograms either.
 
so, is the new equation:
f(-1)(1) = (1/2)(0.5)(4)^2 + 0 - (0.5)(9.8)(1) ??
ans: 900 J ?
 
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