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 with the horizontal.
(b) Find the work when the person has chosen a value
of which ensures him the minimum magnitude of the
force.
Answers
Step-by-step explanation:
I hope this may help you
❣️Hi❣️
Here is your answer =>
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.
Hope it Helps.‼️