JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 10)
_2nd_September_Morning_Slot_en_10_1.png)
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $$\theta $$ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction $$\mu $$. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by $$\mu $$ = ktan $$\theta $$ . The value of k is _________.
Answer
3
Explanation
_2nd_September_Morning_Slot_en_10_3.png)
Apply work energy theorem
Wg + Wf = $$\Delta $$kE
mgsin$$\theta $$ (AC + 2AC) – $$\mu $$mg cos$$\theta $$ $$ \times $$ AC = 0 - 0
$$ \Rightarrow $$ $$\mu $$ = 3tan $$\theta $$
Comments (0)
