Question

A mass of m kg is suspended by a weightless string. the horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction

Answer 1

The work done by the force will be equal to the potentail energy gained by the mass.
Let L be the length of the string ,
F.d= MgL (1 -cos45 )
The horizontal displacement of the mass is d= L sin 45
F L sin45 = Mg L(1 - 45)
F = Mg (sq. root 2 - 1)

USE CONSERVATION OF ENERGY

Let x be the force required
T Sin 45=M kg …....1
T Cos45=x ….......2
Divide equation 1 and 2,
1=M/x
x=M

Answer 2

Work done by trension -Work done by force =Work done by gravitational force

0+F×AB=Mg×AC0+F×AB=Mg×AC

F=Mg(ACAB)F=Mg(ACAB)

AB=lsin45AB=lsin⁡45

=l2–√=l2

AC=OC−OAAC=OC−OA

=l−lcos45=l−lcos⁡45

=Mg(1−1212√)=Mg(1−1212)

AC=OC−OAAC=OC−OA

=l−lcos45=l−lcos⁡45 (l-length of string)

Therefore F=Mg(2–√−1)