Laws Of Motion Question 402
A particle is projected on a rough horizontal ground along positive x-axis from x = 0, with an initial speed of $ v_0 $ The friction coefficient to the ground varies with x as $ \mu(x) = Kx $ Here 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_{\text{r}}}{\sqrt{2Kg}} $
D) $ \frac{r}{r_{r}}{\sqrt{Kg}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Retardation, $ a=-\mu g=-Kx $
or$ V.\left( \frac{dV}{dx} \right)=-Kxg
or$ VdV=-Kgx,dx
or $ \int _{v _{0}}^{0}{VdV}=-Kg\int _{0}^{x}{xdx} $
or $ x=\frac{V _{0}}{\sqrt{g}} $
BETA
