Question #127126

a 20 kg box is initially at rest with a static and kinetic friction of 0.40 is pushed with a force of 70 For 120 s. What is the work done?


1
Expert's answer
2020-07-23T08:46:14-0400

W=FsW=F\cdot s


Ffr=μmg=0.4209.81=78.48N>70NF_{fr}=\mu mg=0.4\cdot 20\cdot 9.81=78.48N>70N


This force (70N70 N) will not be able to move the box. So,

s=0W=0s=0\to W=0 Answer.


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