A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force.
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