a problem for 11 std students (easy)

mass of particle "m"

hey this is veery easy basic ! no tries!! :O

sorry one condition F>μmg

Initial motion... F and friction oppose motion. So,
a=\frac{-(F+\mu mg)}{m}

Now, when it stops v=0 so, and u=v
so,v = \frac{-(F+\mu mg)}{m}t .. or .. t=\frac{mv}{F+\mu mg}
v^2 = 2as.. or.. s=v^2/2a = \frac{mv^2}{2(F+\mu mg)}
For reverse motion,
F acts in the direction of motion and friction opposes. So
a=(F-\mu mg)/m
s=\frac{mv^2}{2(F+\mu mg)} = \frac{(F-\mu mg)t^2}{2m}

t=\frac{mv}{\sqrt{F^2-(\mu mg^2)}}

So total time = t+t

yes ur right ashish

hey just realised that posted this one in the wrong forum :(