Radial Track, Find Loss of TME

bigman8424

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 ??

Related Introductory Physics Homework Help News on Phys.org

Nylex

Energy isn't measured in newtons and you haven't converted your mass to kilograms either.

bigman8424

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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