A block of mass 1 kg start moving at t = 0 with speed `2 m//s` on rough horizontal surface with coefficient of friction 0.2. A horizontal force F is applied in the direction of velocity which varies with time as shown in figure. Find the speed (in m/s) of particle at t = 3 s.

Answer

3

Explanation

`T sin 60^(@) =(mv^(2))/(l sin60^(@))" and "Tcos 60^(@)=mg`
`tan 60^(@) = (v^(2))/(lg sin 60^(@))impliessqrt(3)= (2v^(2))/(lgsqrt(3))`
`v^(2)=(3)/(2)lg`

Now in `2^(nd)` case `(m(2v)^(2))/(lsin60^(@))=4sqrt(3)mg`
So the forces are
`(T+N)cos30^(@)=4sqrt(3)mg`
`(T-N)sin30^(@)=mg`
`implies2N=(4sqrt(3)mg)/(cos30^(@))-(mg)/(sin30^(@))`
`=8mg-2mg=6mg`
`N=3mg`