a) zero

b) 6.24 N

c) 4.34 N

d) 2.68 N

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Let F be maximum force required .

Normal force $N=F+mg \cos \theta$

Since the block does not slip

$mg \sin \theta=\mu N=f$

$\qquad=\mu(F+mg \cos 30^{\circ})$

or $F=\large\frac{mg \sin 30}{\mu}$$-mg \cos 30$

$\qquad=\large\frac{2(10)(1/2)}{.5}$$-2(10)\bigg(\large\frac{\sqrt 3}{2}\bigg)$

$\qquad=20-17.32$

$\qquad=2.68 N$

Hence d is the correct answer

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