Coefficient of kinetic friction and Energy

AI Thread Summary
To solve for the coefficient of kinetic friction, the user has already determined the speed of the block at the bottom of the ramp to be 5.60 m/s. The challenge lies in calculating the frictional force and the work done by friction as the block slides along the floor. The user recognizes that the work done by friction over the distance of 4.5 m must equal the change in mechanical energy of the block. They seek clarification on how to simultaneously find the frictional force and apply it to the work-energy equation. Understanding the relationship between work done by friction and the block's energy loss is crucial for solving the problem.
Leid_X09
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Homework Statement



Starting from rest, a 10.6 kg block slides 3.20 m down to the bottom of a frictionless ramp inclined 30.0° from the floor. The block then slides an additional 4.50 m along the floor before coming to a stop.
(a) Determine the speed of the block at the bottom of the ramp.
(b) Determine the coefficient of kinetic friction between block and floor.
(c) Determine the mechanical energy lost due to friction.


I already solved a) for 5.60 m/s, which is the correct answer. I'm having issues with b, because I know that the equation should be -Fk(4.5m) + 1/2mv^2 = W but I don't know how to apply this equation to find the frictional force. How am i supposed to calculate the work done without finding the frictional force, and vice versa? is there another equation I am not aware of?
 
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-Fk * 4.5m is the work done on the block. It will equal the change in energy of the block as it slides 4.5 m along the floor.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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