1.

A particle of mass m (starting at rest) moves vertically upwards from the surface of earth under an external force overset(to)(F) which varies with height z as overset(to)(F)= (2- alpha z) m overset(to)(g) where alpha is a positive constant. If H is the maximum height to which particle rises. Then

Answer»

`H=(1)/(alpha)`
`H=(2)/(alpha)`
Work done by `OVERSET(to)(F)` during motion UPTO `(H)/(2)` is `(3 mg)/(2 alpha)`
Velocity of particle at `H/2` is `sqrt((g)/(alpha))`

Solution :Resultant force on particle = F - mg
`therefore (mvdv)/(dz) = (1- alpha z) mg rArr (v^2)/(2) = ( z- (alpha z^(2))/( 2) )g`
From maximum height, `v=0`
`rArr z=H= (2)/( alpha) rArr (H)/(2) = (1)/( alpha)`
So, velocity at height `(H)/(2) , v= sqrt(2 (z - ( alpha z^(2) )/( 2) )g)= sqrt(2 ( (1)/(alpha) - (1)/( 2 alpha) ) g) = sqrt((g)/(alpha))`
`therefore` Work done by force F, W `= int_(0)^(1// alpha) (2 -alphaz )` mg dz
`= mg [2Z - (alpha z^(2) )/( 2) ]_(0)^(1//alpha) = mg [(2)/(alpha)- (1)/( 2 alpha) ]= (3 mg)/( 2alpha).`


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