A 50.0 kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.750 and 0.430, respectively. What horizontal pushing force is required to slide the crate across the dock at a constant speed?

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Answer 1 · 2015-10-09T14:55:18

$210,7N$

From Newton's 2nd Law of motion,

$sumF=ma$
$therefore F_(appl)-f_k=ma=0$

Note that acceleration is zero since velocity is constant and by definition $a=(dv)/(dt)$

$therefore F_(appl)=f_k=mu_kN=mu_kmg=0,430xx50xx9,8=210,7N$