Posted by Ivo on Tuesday, October 11, 2011 at 12:25am.
a) (1/2) a t^2, where
a = 1.2 m/s^2
t = 5.0 s
b) Speed at cable breakage = a t
V' = 6 m/s
Let distance travelled after cable breakage be X.
M g X *Uk = (M/2) V'^2
You need the coefficient of kinetic friction Uk. You can get it from the initial cable tension, before it breaks.
350 - M*g*Uk = M*a
Uk = (350 - M a) /(Mg)
= 0.163
Now solve for X.
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