A box weighing 20

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. W_vertical=0 Joules

⇒Total work done on the box = W = W_horizontal+W_vertical

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

⇒ μcosθ_min = sinθ_min

⇒ tanθ_min = μ

⇒ θ_min = tan^-1μ