Abox 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 8 with the horizontal.
(b) Find the work when the person has chosen a value
of 0 which ensures him the minimum magnitude of the
force.
Answers
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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.
Answer:
Verified Answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but Verified Answers are the finest of the finest.
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.