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 is
Will mark the answer as BRAINLIEST!!! if correct

Answer 1

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

Answer 2

Work done by trension -Work done by force =Work done by gravitational force
0+F×AB=Mg×AC
F=Mg(ACAB)
AB=lsin45
=l2–√
AC=OC−OA
=l−lcos45
=Mg(1−1212√)
AC=OC−OA
=l−lcos45 (l-length of string)
Therefore F=Mg(2–√−1)