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Question

A box weighing 2000 N is to be slowly slid through 20 m on a straight track with 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.

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Solution

Given:
Weight=2000 N, s=20 m, μ=0.2

The free-body diagram for the box is shown below:




(a) From the figure,
R+P sin θ-2000=0 ... (i)P cos θ-0.2 R=0 ... (ii)

From (i) and (ii),
P cos θ-0.2 2000-P sin θ=0P cos θ+0.2 sin θ=400P=400cos θ+0.2 sin θ ... (iii)

So, work done by the person,

W=PS cos θ=8000 cos θcos θ+0.2 sin θ=80001+0.2 tan θ=400005+tan θ ...(iv)

(b) For minimum magnitude of force from equation (iii),
ddk cos θ+0.2 sin θ=0 tan θ=0.2
Putting the value in equation (iv),
W=400005+tan θ=400005+0.27692 J

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