Find the least pulling force which acting at an angle of `45^(@)` with the horizontal will slide a body weighing `5 kg` along a rough horizontal surface . The coefficient of friction `mu_(s) = mu_(k) = 1//3` If a force of doude this is applied along the same direction , find the resulting acceleration of the block

Explanation

When the block is about to start sliding the friction force acting on block reaches to its limiting value `f = mu_(s) N`

In verticle direction: ` N + P sin`
` 45^(@) = mg `
`rArr N = mg - P sin 45^(@)`
In horizontal direction:
`P cos 45^(@)= mu_(s)N`
or `P cos 45^(@)= mu_(s)(mg - P sin 45^(@))`
or `P = (mu_(s)(mg))/(cos 45^(@) + mu_(s) sin 45^(@)) = (25)/(sqrt(2))N`
If applied force is`2p` : Friction will be kinetic nature . The block will move ` f = mu_(k)N` .
`N = mg - 2P sin 45^(@)` and `2P cos 45^(@)- mu_(s)N = ma`
`2P cos 45^(@) - mu_(s)(mg - 2P sin 45^(@)) = ma`
`a = (2P)/(msqrt(2)) (mu_(k) + 1 ) - mu_(k)g = (10)/(3)ms^(-2)`