Question

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

Answer 1

m/√2 will be the anwer

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)