Coefficient of kinetic friction and pulling force

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To determine the coefficient of kinetic friction for a 20-kilogram sled being pulled at a constant velocity with an 80 Newton force at a 30-degree angle, the horizontal component of the pulling force must equal the kinetic friction force. The normal force can be calculated by subtracting the vertical component of the pulling force from the sled's weight. The formula for kinetic friction is the frictional force divided by the normal force. Since the sled moves at constant velocity, the net force is zero, indicating that the pulling force's horizontal component balances the frictional force. Understanding these relationships is crucial for calculating the coefficient of kinetic friction accurately.
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hi everybody.. can i ask how to get a coefficient of kinetic friction if a 20 kilogram sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80 Newton and is directed at an angle of 30 degrees above the horizontal. please answer my question.. i badly need you help..
 
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I assume you know the formula for finding kinetic friction. If the sled is moving at constant velocity, what will be the relationship between the horizontal component of the pulling force and the kinetic friction? How do you find the normal force given the weight of the sled and the upward force that the person is helping ease the contact between the sled and 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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