A block of mass m resting on a smooth horizontal plane starts moving due to a constant force F=mg//3 of constantmagnitude. In the process of its rectilinear motion the angle theta varies as theta=lamdas, where lamda is constant and s is the distance traversed by the block from its intial position. Find the velocity of the block as a function of the angle theta. Also show that the block will never lose contact with the ground.

A block of mass m resting on a smooth horizontal plane starts moving d

1.	A block of mass m resting on a smooth horizontal plane starts moving due to a constant force F=mg//3 of constantmagnitude. In the process of its rectilinear motion the angle theta varies as theta=lamdas, where lamda is constant and s is the distance traversed by the block from its intial position. Find the velocity of the block as a function of the angle theta. Also show that the block will never lose contact with the ground.
Answer» Solution : `x-`direction`:` `Fcostheta=ma` `a=(F)/(m)costheta=(mgcos(lamdas))/(3m)` `v(dv)/(dx)=(g)/(3)COS(lamdas)` `int_(0)^(v)vdv=(8)/(3)int_(0)^(s)cos(lamdas)ds` `\|(v^(2))/(2)\|_(0)^(v)=(g)/(3)(\|sin(lamdas)\|_(0)^(s))/(lamda)` `v^(2)=(2g)/(3lamda)sin(lamdas)` `v=sqrt(2gsintheta)/(3lamda)` `y-`direction`:` `N+Fsintheta=mg` `N=mg-(mg)/(3)sintheta` `(sintheta)_(max)=1,N_(MIN)=(2mg)/(3)` Since normal reaction will NEVER will never be zero and hence the block will never LOSE constact with the ground.