Question

FP 30.0° ÈN The 10.0-kg box is now pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction remains at 0.30. Calculate its acceleration. What would be the coefficient of static friction? Ffr mg

Answer 1

Ipsimo 90 3 boy mg the ban then toa my 90 x 6 2 Forces acting on box fe Fp at om angle 30° impen haricental To cose Foro Normat Porco ben Errettono fonce For ма of ban meign of There is no mottom of the bom along y apors so we com write Fosimo f No mg N2 ling. Epsime rf a be one acceleration or Mar/Fp Coser Fr Fiora frictional fores ar kinetic Freil eng - Fosma) from Equation ③ we can write Fp Cosor u cong-Fosna) PM 404 Cos(50) 0.3 %0 (1080908 40 % Simloo) 0.31 (98-10143) Or, loa z 20 NBC 0.3 y (98-20) any loa 2003- 23.4 ONA loa Oy 10124 myse acceleration of box = 1.12 m2 Static friction s is the priction that exists between a stationary abject and the surface on which it is resting. f.ee acceleration of the box to so we can write If us best Bocorro Fp Cose = F ON, Ip Coser useng - Fpština) Focoso mga Fp Sima) 40 x Cos (30°) 10*9.8 - 40% Sim 30) 40 * 98- 20 coefficient of stalre forchom = 0.44 Static friction - MSN us les = us 0.44 Scanned with CamScanner