A 2-kg block sits on an incline where the top part of the incline has a coeffici

Question

                    A 2-kg block sits on an incline where the top part of the incline has a coefficient of kinetic friction (mk) of 0.70. The bottom section of the plane has mk= 0.8. The angle of inclination is 40 degrees. The block is released and travels 10 m along                     the initial part of the incline and then enters the lower section. Calculate how far the block travels along the second section before it is brought to a                     stop.                 

                    
                

                                     

A 2-kg block sits on an incline where the top part of the incline has a coefficient of kinetic friction (mk) of 0.70. The bottom section of the plane has mk = 0.8. The angle of inclination is 40 degrees. The block is released and travels 10 m along the initial part of the incline and then enters the lower section. Calculate how far the block travels along the second section before it is brought to a stop.

Explanation / Answer

in first section negative accerlation of block = mgsin(@) - umgcos(@) = 20 sin40 - 0.7*20*cos40 = 2.13

velocity of block when block goes 10 m in first section = > v^2 = u^2 + 2as = V^2 = 0 + 2*2.13*10

velocity of block at the starting of second section = 2*2.13*10

deaccerlation of block in second section => a = 20 sin40 - 16 cos40 = .59

so the distance covered by block = v^2 = u^2 + 2as => 0 = 2*2.13*10 - 2*.59 *S

so distance S= 3.65m

thank u

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