A block of mass m slides down the plane inclined at an angle `60^@` with an acceleration g/2. Find the co-efficient of kinetic friction.

Explanation

Kinetic friction comes to play as the block is moving on the surface. The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction `f_k` along the smface. · Along the x-direction
`mg sin theta - f_k = ma `
But a g/2
`mg sin 60^@ - f_k = mg//2`
` (sqrt3)/(2) mg - f_k = mg//2`
` f_k = mg ( (sqrt3)/(2) - 1/2)`
` f_k = ( (sqrt3 -1)/(2) ) mg `
There is no motion along they-direction as normal force is exactly balanced by the ` mg cos theta `
` mg cos theta = N = mg//2`
` f_k =mu_k N = mu_k mg//2`
`mu_k = ( (sqrt3 - 1)/(2) mg )/( (mg)/(2) )`
` mu_k = sqrt3 - 1`