d = 0.5a*t^2 = 10 m.
0.5a*(5.5)^2 = 10
15.125a = 10
a = 0.661 m/s^2.
V = a*t = 0.661 * 5.5 = 3.636 m/s.
d = Vo*t - o.5a*t^2
d=3.636*5.10 - 0.5*0.661*5.10^2=9.95 m.
A dockworker applies a constant horizontal force of 71.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 10.0m in a time of 5.50s .
If the worker stops pushing after 5.50s , how far does the block move in the next 5.10s ?
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2 answers