Radial Track, Find Loss of TME

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SUMMARY

The discussion focuses on calculating the loss of Total Mechanical Energy (TME) for a 500g block sliding 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 end. The formula used is TMEf - TMEo, where the final kinetic energy (K.E.f) and potential energy (P.E.f) are compared to the initial kinetic energy (K.E.o) and potential energy (P.E.o). The correct calculation shows that the loss of TME is 900 J, confirming that energy is measured in joules, not newtons, and that mass must be converted to kilograms for accurate results.

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bigman8424
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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 ?
 
Last edited:

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