1.

At t=0, a force F=at^(2) is applied to a small body of mass m at an angle alpha resting on a smooth horizontal plane

Answer»

velocity of the body at the moment it brakes off the plane is `sqrt((mg^(3))/(9a tan^(2)alphasinalpha))`
the distance travelled by the body before breaking off the plane is `(mg^(2))/(12a sinalphatanalpha)`
Its acceleration at the time of breaking off the plane is `g cot ALPHA`.
Time at which it breaks off the plane is `sqrt((mg)/(asinalpha))`.

Solution :For breaking off the plane
`F sin alpha=mg`
`rArrat_(0)^(2)sinalpha=mg`
`RARR t_(0)=sqrt((mg)/(asinalpha))`
SPEED at time of breaking off
`v=int_(0)^(t_(0))(at^(2)COSALPHA)/(m)dt=(at_(0)^(2)cosalpha)/(3m)`
`=(acosalpha)/(3m)*(mg)/(asinalpha)sqrt((mg)/(asinalpha))`
`=sqrt((mg^(3))/(9atan^(2)alphasinalpha))`
`a=(Fcosalpha)/(m)=(at_(0)^(2)cosalpha)/(m)=(AMG)/(asinalpha)*(cosalpha)/(m)`
`=gcotalpha`
`s=intvdt=int_(0)^(t_(0))(at^(3))/(3m)cosalphadt`
`=(a)/(12m)t_(0)^(4)=(a)/(12m)*(m^(2)g^(2)cosalpha)/(a^(2)sin^(2)alpha)`
`=(mg^(2))/(12a tanalphasinalpha)`


Discussion

No Comment Found

Related InterviewSolutions