Answer :

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Where, w=mg=weight of the box=2000 N


g=acceleration due to gravity=10ms-2


F=force of pull being applied


N=normal reaction


Force of friction=μN


According to the free body diagram of the box,


N=w-Fsinθ


Fcosθ=μN


Fcosθ=μ(w-Fsinθ)


F(cosθ+μsinθ) =μmg



(a.) The horizontal component of work done on the box by pulling is the product of the horizontal component of force applied on the box and the resulting horizontal displacement of the box






The vertical component of work done will be zero as there is no displacement in the vertical direction.


i.e. Wvertical=0 Joules


Total work done on the box = W = Whorizontal+Wvertical




(b.) Letθmin be the angle for which F is minimum. We will use the first derivative method here.




μcosθmin = sinθmin


tanθmin = μ


θmin = tan-1μ


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