Question

A box weighing $2000N$ is to be slowly slid through $20m$ 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 $\theta$ with the horizontal. (b) Find the work when the person has chosen a value of $\theta$ which ensures him the minimum magnitude of the force.

Answer 1

According to the given conditions :-

Let us first draw the diagram :- Refer to the attachment

Here weight of the box mg = 2000 N

μ = 0.2

distance s = 20 m.

.

so if the pull is applied bt the person is F .

Then the Normal force, N +F.sinθ = W { = 2000 N}

N = 2000 - Fsinθ

Force of friction is = μN

= μ (2000- F sinθ )

= 0.2 (2000- F sinθ)

=400 -0.2 F sinθ .

According to the given conditions driving force is equal to the frictional force.

so, F.cosθ = 400 - 0.2 F.sinθ

F (cosθ + 0.2sinθ) = 400

multiply both sides by 5 then,

F (5cosθ + sinθ) = 2000

F = 2000/ [(5cosθ + sinθ)]

(a) Now for work done

work done by the person pulling the box is

= F cosθ × 20 J

= 2000/[(5cosθ + sinθ)] cosθ × 20 J

W=40000 / (5 + tanθ) J. -------(i)

Hence the work done by the person is 40000 / (5 +tanθ) J.

(b) For minimizing the F, df/dθ = 0

-2000(-5sinθ+cosθ) / (5cosθ + sinθ)² =0

hence, cosθ =5sinθ

sinθ/cosθ = 1/5

tanθ =1/5

so by putting this tanθ value in equation (i), work done expression.

W = 40000/ (5 + 1/5) J

W = 7692 J.