The figure shows a 0-41-kg block sliding from A to B along a frictionl.docx

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The figure shows a 0.41-kg block sliding from A to B along a frictionless surface. When the block reaches B, it continues to slide along the horizontal surface BC where the kinetic frictional force acts. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A is 37 J, and the heights of A and B are 12.0 and 7.0 m above the ground, respectively. A. How fast is the block moving when it reaches B? B. Find the coefficient of kinetic friction between the horizontal surface BC and the block if the distance between points B and C is 7.5 m. Note: The distance between points B and C is not drawn to scale. A-12.0m Solution A) the problem can be solved from energy conservation Potential energy of block at A=0.41*9.8*12=48.216 J Kinetic energy at A=37 J Potential energy at B=0.41*9.8*7=28.126 J Kinetic energy at B=48.216+37-28.126=57.09 J Velocity of block at B=sqrt(57.09*2/0.41)=16.69 m/s b) let the deceleration be d d=16.69Â²/(2*7.5)=18.57 m/sÂ² Coefficient of kinetic friction=18.57/9.8=1.89 .

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The figure shows a 0-41-kg block sliding from A to B along a frictionl.docx

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