Derive an expression work done in sliding body down a rough inclined plane-

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Toppr Admin · Accepted Answer

Friction force- f-x3BC-R-x3BC-mgcos-x3B8-x3BC- is the coefficient of the friction-xA0-xA0-Net force in forward direction- F-mgsin-x3B8-x2212-x3BC-mgcos-x3B8-Let the distance travelled by the body is -S-work done- W-F-S-xA0-W-mgsin-x3B8-x2212-x3BC-mgcos-x3B8-S

Derive an expression for work done in sliding body down a rough inclined plane.

friction force, $f = μ R = μ m g c o s θ$

$μ$ is the coefficient of the friction.

Net force in forward direction, $F = m g s i n θ - μ m g c o s θ$

Let the distance travelled by the body is , $S$ .

work done, $W = F . S$

$W = (m g s i n θ - μ m g c o s θ) . S$

Derive an expression for work done in sliding body down a rough inclined plane.

friction force, f=μR=μmgcosθμ is the coefficient of the friction. Net force in forward direction, F=mgsinθ−μmgcosθLet the distance travelled by the body is ,S.work done, W=F.S W=(mgsinθ−μmgcosθ).S

friction force, $f = μ R = μ m g c o s θ$

$μ$ is the coefficient of the friction.

Net force in forward direction, $F = m g s i n θ - μ m g c o s θ$

Let the distance travelled by the body is , $S$ .

work done, $W = F . S$

$W = (m g s i n θ - μ m g c o s θ) . S$