Question

A small block when released on an inclined plane,
it first slides down and then stops after sliding
down a height 'h'. This strange behavior is due to
the coefficient of friction that is hear proportional
to the distance slid by the block. The maximum
gh
speed of the block during this motion is Find
'n' value(Acceleration due to gravity is 'g')​

Answer 1

The maximum speed of the block during the motion is

$\boxed{\sqrt{\frac{gh}{2}}}$

Explanation:

The coefficient of friction is proportional to the distance slid by the block

Let the distance slid is x and coefficient of friction $\mu$

Then

$\mu\propto x$

$\implies \mu=kx$

Let the plane is inclined at angle $\theta$

If the total distance covered on inclined plane is s

Then

$\sin\theta=\frac{h}{s}$

Net force while sliding

$F=mg\sin\theta-\mu mg\cos\theta$

$a=\frac{F}{m}$

$\implies a=g\sin\theta-kxg\cos\theta$

$\implies v\frac{dv}{dx}=g(\sin\theta-kx\cos\theta)$

$\implies vdv=g(\sin\theta-kx\cos\theta)dx$

$\implies\int_0^v vdv=g\sin\theta\int_0^x dx-kg\cos\theta\int_0^x xdx$

$\implies \frac{v^2}{2}=gx\sin\theta-kg\frac{x^2}{2}\cos\theta$

At distance s, v = 0

Thus,

$s\sin\theta=k\frac{s^2}{2}\cos\theta$

$\implies k=\frac{2\tan\theta}{s}$

For max v

$\frac{dv^2}{dx}=0$

$\implies g\sin\theta-kgx\cos\theta=0$

$\implies x=\frac{\tan\theta}{k}$

Also

$\frac{d^2v^2}{dx^2}=-kg\cos\theta$    which is negative

Thus, the speed will be maximum

Maximum speed

$\frac{v_{max}^2}{2}=g(\frac{\tan\theta}{k}\times \sin\theta-k\frac{1}{2}(\frac{\tan\theta}{k})^2\cos\theta)$

$\implies \frac{v_{max}^2}{2}=\frac{g\tan\theta}{k}(\sin\theta-\frac{\tan\theta.\cos\theta}{2})$

$\implies \frac{v_{max}^2}{2}=\frac{gs}{2}(\sin\theta-\frac{\sin\theta}{2})$

$\implies \frac{v_{max}^2}{2}=\frac{gs\sin\theta}{4}$

$\implies v_{max}^2=\frac{gs}{2}\times\frac{h}{s}$

$\implies v_{max}^2=\frac{gh}{2}$

$\implies v_{max}=\sqrt{\frac{gh}{2}}$

Hope this answer is helpful.

Know More:

Q: A small block is released from rest from the top of a rough inclined plane of height 1 m and inclination of 37°. If it takes 1 second to reach the bottom, the coefficient of kinetic friction on the plane and the velocity with which the block reaches the bottom (in m/s) are respectively.

Click Here: https://brainly.in/question/7573293