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In the figure a fastidious worker pushes directly along the handle of a mop with a force....

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- ?Lv 72 months agoFavorite Answer
a)

(m*g+F*sin 45)*μ k = F*cos 45

0.70*9.806*0.36 = F*0.707*(1-0.36)

F = 0.70*9.806*0.36 / (0.707*(1-0.36)) = 5.46 N

b)

(m*g+5,46*y)*μ s = 5.46*x

0.49*9.806 = 5.46(x-0.7y)

4.80/5.46 = x-0.7y

0.880 = x-0.7 y

x^2+y^2 = 1

0.880 = √ (1-y^2)-0.7y

0.880^2+0.49y^2+1.232y = 1-y^2

-1.49y^2-1.232y+0.226 = 0

y = (1.232-√1.232^2+1.346)/-2.98 = 0.154 = sen (90-Θ ')

x = √1-0.154^2 = 0.988 = cos (90 -Θ ')

90-Θ ' = arctan y/x = 8.88°

Θ ' = 81.12°

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