A block is sliding down a rough inclined plane of inclination 30^ with a constant acceleration of 2.5 m s ⁻² . If g = 10 m s ⁻² , the coefficient of kinetic friction between the block and the incline is:

Step-by-step solution

The equation of motion for a block sliding down a rough inclined plane is given by: mg - f_k = ma where f_k = _k N = _k mg . Substituting the expression for kinetic friction: mg - _k mg = ma Dividing the entire equation by m : g - _k g = a Substitute the given values ( g = 10 m s ⁻² , = 30^ , a = 2.5 m s ⁻² ): 10 30^ - _k (10) 30^ = 2.5 10 ( 1 2 ) - _k (10) ( 3 2 ) = 2.5 5 - 5 3 _k = 2.5 5 3 _k = 5 - 2.5 = 2.5 _k = 2.5 5 3 = 1 2 3 Answer: 1 2 3