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Pushing a box (friction)

  • Thread starter Chandasouk
  • Start date
  • #1
165
0

Homework Statement



A stockroom worker pushes a box with mass 11.1 kg on a horizontal surface with a constant speed of 3.80 m/s. The coefficient of kinetic friction between the box and the surface is 0.170.


Homework Equations


F=ma
w=mg


What horizontal force must be applied by the worker to maintain the motion?

I found that the Fapplied is 18.5N


If the force calculated in part A is removed, how far does the box slide before coming to rest?

This question I'm having problems on. Here is what I did

18.5N = 11.1kga

a=1.66666667 m/s^2

Then I used the equation of

Vfinal^2=Vinitial^2+2a[tex]\Delta[/tex]x

[tex]\Delta[/tex]x = 0 - 14.44m/s
--------------
2(1.66666667)


[tex]\Delta[/tex]x=4.3m ?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Yes that is correct.

Also you could have used conservation of energy.

Note that your a=1.66666667 m/s2 should really be a=-1.66666667 m/s2

But I believe you put it as negative while working it out, so just remember to write the - sign else your teacher might give you it wrong.
 

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