Question

block of mass M is released on the top of a smooth inclined plane of length X and inclination theta as shown in figure.
horizontal surface is rough if block comes to rest after moving a distance d on the horizontal surface the coefficient of friction between block and surface is ​

Answer 1

Answer:

Mechanical energy is conserved

u=0, v=0

mgh-(M)mgd=Kf - Ki

=(1/2)mv² -(1/2)mu²

=0

So, mgh=Mmgd

M=h/d...... But, we know that h=xsin(theta)

Thus, M=xsin(theta)/d

Where, M=co-efficient of friction

Answer 2

Answer:

M=xsinθ/d.

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.