1.

A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until string makes an angle of 45^@ with the initial vertical direction is:

Answer»

`(Mg)/(sqrt2)`
`Mg(sqrt(2)- 1)`
`Mg (sqrt(2) + 1)`
`Mg sqrt(2)`

Solution :According to work energy theorem “work done by all forces is equal to change kinetic energy”.
`:. W_("total") = Delta K`
`W_(F) + W_("gravity") + W_(T) = (K_("FINAL") - K_("INITIAL")) = (0-0) = 0`
Work done by `F , W_(F) = F xx l SIN 45^@ = (F1)/(sqrt(2))`
Work done by gravity ,
`W_("gravity") = -Mg(1 - l cos 45^@) = - M gl(1 - 1/(sqrt2))`
Work done by TENSION = 0
`:. (F1)/(sqrt2) + (- (Mgl(sqrt(2) - 1))/(sqrt(2))) = 0 "or" F = Mg(sqrt(2) - 1)` .


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