IIT JEE , AIEEE Forum - Physics , Mechanics

zodvirus

1) A mass of M kg is suspended by a weight less string. the horizontal force required to displace it until the string makes an angle of 45 degree with initial vertical direction ? options a)Mg/

2 b)Mg/ (

2-1 ) c) Mg/ (

2+1 ) d)Mg

2

govindsuku

the answer is c

govindsuku

the block has sum disp in vertical as well as horizontal direction
vertical disp = l - lcos45
and horizontal disp = lsin45

work done by horizontal force = increase in potential energy

so
F x hor. disp = mg x vert elevation
Fx l/root(r) = mgl ( root(2)-1)/ root(2)
this gives
F = mg( root (2) - 1 )

Multiply and divide by ( root (2) + 1 )

and we will get ans as c

its_my_life_001

answer is option C

sooraj

yep. itz c

edison

Let L be length of string.

The work done in displacing the mass is given by FLsin 45.

This work is stored as gravitational potential energy.

The mass rises from a length L to Lcos45.

Hence change in gravitational potential energy is Mg (L- L cos 45). Using work- energy theorem,

FLsin 45 = Mg (L - L cos 45)

F= MG(1-cos45)/sin45

F = Mg (

2 - 1)

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