On a rough horizontal x-y plane a body starts moving from origin with initial velocity `(V_(0))hat(i)`. A constant force `(F_(0))hat(j)` is always acting on the body. Kinetic friction on the body has magnitude `F_(0)`. Find the magnitude (in m/s) of velocity of the particle when its direction makes an angle of `60^(@)` from y-axis. `V_(0) = 12 m//s`, `F_(0)` = 10N, mass of body = 1kg.

Answer

1

Explanation

Till rolling starts kinetic friction acts on the bodies.
F.B.D. of sphere fro no vertical acceleration N = mg
`implies f_(k)=mumg("as f"_(k)=mu N)`
`implies a=mug`
`v=mu g t`
Also `alpha=(mumgr)/(I)impliesomega=(mum g t)/(I)`
The velocity of the lower most point of the sphere at time t is
`v_(s)=v+omegar`
`v_(s)= mu g t + (mu mg r^(2))/(I)t`
The F.B.D. of the plank
`implies a_(p)=-((F+f_(k))/(M))=-((F+mumg)/(M))`
`v_(p)=v_(0)-((F+mumg)/(M))t`
`V_(S)=V_(P)implies t=(v_(0))/([(7)/(2)mug+((F+mumg)/(M))])`