Ice block sliding distance after pushing force stops

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Homework Statement



A dockworker applies a constant horizontal force of 83.0 to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 10.0 in a time 5.30 .

If the worker stops pushing at the end of 5.30 , how far does the block move in the next 4.20?

Homework Equations



d=v(i)t+.5at^2
a=(vf-vi)/t

The Attempt at a Solution



I solved Vi by dividing distance/time, and then solved for acceleration (-.449235) by dividing (Vf-Vo)/t with Vf=0 and Vi=1.88. I used 4.20 as "t". I plugged the results into d=v(i)t+.5at^2 and got 3.96 as an answer, however masteringphysics said I was wrong. What am I doing wrong??
 
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vi stands for vinitial which is ZERO in this case as the block starts from rest. Given that you can solve for a and then the velocity at 5.30 secs