Asked by Karina
A dockworker applies a constant horizontal force of 90.0Nto a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.0min a time of 5.20s.If the worker stops pushing after 5.20s, how far does the block move in the next 5.30s
Answers
Answered by
Henry
Vo = 12m/5.2s = 2.31 m/s.
Vf = 0.
a=(Vf-Vo)/t=(0-2.31)/5.30=-0.436 m/s^2.
d = Vo*t + 0.5a*t^2.
d = 2.31*5.3 - 0.218*(5.3)^2 = 6.1 m.
Vf = 0.
a=(Vf-Vo)/t=(0-2.31)/5.30=-0.436 m/s^2.
d = Vo*t + 0.5a*t^2.
d = 2.31*5.3 - 0.218*(5.3)^2 = 6.1 m.