A block of mass $M$ slides down a rough inclined plane of height $h$ through a slope distance $S$ . If F be the force of friction, then the kinetic energy at the bottom will be

M g h

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M g S

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M g h + F S

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M g h - F S

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Solution

The correct option is D $M g h - F S$
From law of conservation of energy,
T.E. remains constant
$∴ P . E . = K . E . + F . S$
$∴ K . E . = P . E . - F . S = M g h - F . S$

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A block of mass M slides down a rough inclined plane of height h through a slope distance S. If F be the force of friction, then the kinetic energy at the bottom will be

The correct option is D Mgh−FSFrom law of conservation of energy,T.E. remains constant∴P.E.=K.E.+F.S∴K.E.=P.E.−F.S=Mgh−F.S

A block of mass $M$ slides down a rough inclined plane of height $h$ through a slope distance $S$ . If F be the force of friction, then the kinetic energy at the bottom will be

The correct option is D $M g h - F S$
From law of conservation of energy,
T.E. remains constant
$∴ P . E . = K . E . + F . S$
$∴ K . E . = P . E . - F . S = M g h - F . S$