Stockroom worker pushes a box homework problem

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moondawg
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Homework Statement



A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.50m/s. The coefficient of kinetic friction between the box and the surface is .20. What horizontal force must the worker apply to maintain the motion?

Homework Equations



F=ma
F=[tex]\mu[/tex]N


The Attempt at a Solution


First I drew my free body diagrams of course and found my Normal force to be 109.76N. Then i used this number and plugges it into F=ma(F=11.2x3.5) and got 39.2. Then using the 2nd equation above I found F=21.95. I was not sure if i was doing it correctly but then i added the 39.2+21.95= 61.152 as my F and answer. But I realized this cannot be right because i used 3.5 as my acceleration when it is just a constant velocity so I do not know how to even approach this question... PLEASE HELP!
 
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If velocity is constant than all forces are equal. Force of friction will equal the applied force. Just do Force friction= normal fprce x coeddicient of kinetic friction. Since forces are equal they will be the same value.