Asked by Anonymous
A carton on a ramp that has an angle of 25 degrees. If coefficient of kinetic friction is .12, what is the speed at the bottom of the ramp, knowing that the carton was 9.3m from the base?
Answers
Answered by
Henry
h = 9.3sin25 = 3.93m.
Vf^2 = Vo^2 + 2g*h,
Vf^2 = 0 + 19.6*3.93 = 77.03,
Vf = 8.78m/s
Vf^2 = Vo^2 + 2g*h,
Vf^2 = 0 + 19.6*3.93 = 77.03,
Vf = 8.78m/s
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