A small block of mass m is projected horizont | vedclass

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(A) The normal force acting on the block is $N = mg$.The frictional force is $f = \mu N = (cx)mg$.Using Newton's second law,the retardation $a$ is giv

Vedclass · Accepted Answer

$\frac{u}{\sqrt{cg}}$

Vedclass · Answer

(A) The normal force acting on the block is $N = mg$.The frictional force is $f = \mu N = (cx)mg$.Using Newton's second law,the retardation $a$ is given by $a = \frac{f}{m} = \frac{(cx)mg}{m} = cgx$.Since the block is decelerating,$a = v \frac{dv}{dx} = -cgx$.Separating the variables and integrating from the initial velocity $u$ to final velocity $0$ over the distance $x$:$\int_{u}^{0} v \, dv = \int_{0}^{x} -cgx \, dx$$\left[ \frac{v^2}{2} \right]_{u}^{0} = -cg \left[ \frac{x^2}{2} \right]_{0}^{x}$$0 - \frac{u^2}{2} = -cg \frac{x^2}{2}$$\frac{u^2}{2} = \frac{cgx^2}{2}$$x^2 = \frac{u^2}{cg}$$x = \frac{u}{\sqrt{cg}}$

$A$ small block of mass $m$ is projected horizontally with speed $u$. The coefficient of friction between the block and the plane is given by $\mu = cx$,where $x$ is the displacement of the block on the plane. Find the maximum distance covered by the block.

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