A block having a mass 3kg is initially at rest on a horizontal surface. The coefficient of static friction `mu_(s) =0.3` between theblock and the surface and `mu_(k)` is 0.25 .A constant force F of 50 N, acts on the body at the angle `theta = 37^(@)`. What is the acceleration of the block ?

Explanation

We have two possibilities here, the block may remain at rest, or it may accelerate towards the right. The decision hinges on whether or not the x - component of the force F has manitude, less than or greater than the maximum static friction force. The x - component of F is
`F_(x) = F cos theta = (50 N ) ( 0.8) = 40 N `
To find `f_("s, max")`, we first calculate the normal force N, wherther or not the block accelerates horizontally, the sum of the y-component of all the forces on the sum of the y-component of all the forces on the block is zero.
`N - F sin theta - mg = 0`
or `N = F sin theta + mg = ( 50 N ) ( 0.6) + (3 kg ) ( 9.8 ms^(-2))`
`= 59.4N`

The maximum static frictional force
`f_(s)`, max `= mu_( s) N ( 0.3) ( 59.4N) = 17.8 N `
This value is smaller than the x-component of F, hence the block moves. We now interpret the force f in the figure as a kintice frictional force. This value is obtained as
`f_(K) = mu_(k) N = ( 0.25) ( 59.4N) =14.8 N `
Therefore resultant force in the x-direction is
`sumF_(x) = F cos theta - f = 40 N - 14.8 N =25.2N`
Then the acceleration 'a' of the blockis
`a = (25.2N)/( 3 kg ) = 8.4 ms^(-2)`