A body weighing 20 kg just slides down a rough inclined plane that rises 5 in 12. The coefficient of friction is

Options

A. ` 0.46 `
B. `4.6 `
C. ` 0.52 `
D. `0.12 `

Answer

A

Explanation

The given situation can be shown as
As the plane rises 5 in 12
` therefore sin theta = ( 5 ) /( 12 ) `
` (##ARH_EGN_SP_12_E01_048_S01.png" width="80%">
and ` cos theta = sqrt ( 1 - sin ^ 2 theta ) = sqrt ( 1 + ((5 ) /( 12 )) ^ 2 ) `
` = ( sqrt ( 119))/( 12 ) `
So, the coefficient of friction
` mu = tan theta = ( sin theta )/( cos theta ) = ( 5 ) /( 12 ) xx ( 12 )/( sqrt ( 119)) = ( 5 )/( sqrt (119)) = 0.46 `