A long chain of mass of mass `M` length `L` is being pulled with constant velocity on a rough incline with coefficient of friction `mu`. What rate frictional force on chain is increasing

Options

A. `(3)/(10) mu (M)/(L) gv`
B. ` (5)/(6) mu (M)/(L) gv`
C. Zero
D. `mu (M)/(L) gv`

Answer

A

Explanation

In a small time let dx length crossing from lower incline to upper incline
`df = mu (M(dx))/(L) " g cos" 37^(@) - mu (M)/(L) "dx cos" 60^(@)`
`df = mu (M)/(L) g ("cos" 37^(@) - "cos" 60^(@)) "dx"`
`(df)/(dt) = (muM)/(L) g ((4)/(5) - (1)/(2)) (dx)/(dt) = (3)/(10) (mu M)/(L) gv`