1sorry one condition F>μmg
A particle is initially situated at origin . It starts moving with a velocity v towards right .During motion a force F always acts on the body towards left . given coefficient of friction μ , find the time taken by particle to return to origin .
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6 Answers
106Initial 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
