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Radial Track, Find Loss of TME

  • Thread starter bigman8424
  • Start date
25
0
looks like a fun problem, just gotta 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 ??
 
551
1
Energy isn't measured in newtons and you haven't converted your mass to kilograms either.
 
25
0
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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