Question

A mass M kg is suspended by a weightless string. The horizontal force that is required to displace it untill the string makes an angle 45∘ with initial vertical direction is

Answer 1

min horizontal force is Mg

Answer 2

Answer:

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)

Explanation: