Laws Of Motion Question 402

Question: A particle is projected on a rough horizontal ground along positive x-axis from x = 0, with an initial speed of $ {{r} _{r}} $ The friction coefficient to the ground varies with x asHere K is a positive constant. The particle comes to rest at x equal

Options:

A) $ \frac{{{r} _{r}}}{\sqrt{r}} $

B) $ \frac{r{{r} _{r}}}{Kg} $

C) $ \frac{{{r} _{r}}}{\sqrt{2Kg}} $

D) $ \frac{r{{r} _{r}}}{\sqrt{Kg}} $

Show Answer

Answer:

Correct Answer: A

Solution:

Retardation, $ a=-\mu g=-Kx.g $

or$ V.( \frac{dV}{dx} )=-Kxg $

or$ VdV=-Kgx.Dx $

or $ \int _{v _{0}}^{0}{VdV}=-Kg\int _{0}^{x}{xdx} $

or $ x=\frac{V _{0}}{\sqrt{Kg}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक