Question

L 16. A block of mass m is released on the top of a smooth inclined plane of length x and inclination 0 as shown in figure. Horizontal surface is rough. If block comes to rest after moving a distance d on the horizontal surface, then coefficient of friction between block and surface is m х T7 к d x sin e X cose (1) (2) 2d 2d x sine X cosᎾ (3) (4) d d​

Answer 1

Answer:

xsin0

d

Explanation:

tick the correct

Answer 2

Answer:

M=xsinθ/d.

$\huge \dag \mathbb \purple{\underline{EXPLANATION}}$

Since, for the question we get that the initial velocity and the final velocity is zero, which is u=0, v=0. So, we can say the when the block moves from its initial potion to the final position then there is conservation of Mechanical energy.

So, now mgh-μmgd=Kf - Ki ( where μ is the coefficient of friction).

Kf - Ki =(1/2)mv² -(1/2)mu² which will be = 0.

Therefore, mgh=μmgd.

So, on solving we get that μ=h/d. While taking the value of height h we get that the h=xsinθ.

• Thus, M=xsinθ/d.

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