A block of mass m starts at rest at height h on a frictionless inclined plane. block slides down pla

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5.Work, Energy, Power and Collision

hard

A block of mass $m$ starts at rest at height $h$ on a frictionless inclined plane. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction $μ$ , and compresses a spring with force constant $k$ a distance $x$ before momentarily coming to rest. Then the spring extends and the block travels back across the rough surface, sliding up the plane. The block travels a total distance $d$ on rough horizontal surface. The correct expression for the maximum height $h’$ that the block reaches on its return is

A

$mgh’\ =\ mgh\ -\ \mu mgd$

B

$mgh’\ =\ mgh\ +\ \mu mgd$

C

$mgh’\ =\ mgh\ +\ \mu mgd\ +\ kx^2$

D

$mgh’\ =\ mgh\ -\ \mu mgd\ -\ kx^2$

Solution

Final $P.E.$ of block $=$ Initial $P.E.$ of block + work done by friction

$\therefore $ $mgh'\ =\ mgh\ -\ \mu mgd$

Standard 11

Physics

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