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Question

A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angle of 45o with the initial vertical is :

A
Mg(21)
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B
Mg(2+1)
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C
Mg2
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D
Mg2
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Solution

The correct option is B Mg(21)
Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work energy theorem. It is given that body is taken slowly so that its speed doesn't change, then
ΔK.E=0
=WF+WMg+Wtension
[symbols have their usual meanings]
WF=F×lsin450,
WMg=Mg(llcos450)
and Wtension=0 ( tension is always perpendicular to the motion of block)
F×lsin45oMg(llcos45o)+0=0
F=Mg(21)

417303_41093_ans_7207314907b94043944af3ca1aa1ac9f.png

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