Coefficient of kinetic friction?? Please help?

[匿名]

2016-06-20 6:09 am

A block of mass m=2.2kg is placed on an incline with angle= 25 degrees, if the block slides down the incline at a constant speed, what is the coefficient of kinetic friction?

Please help! I will award best answer for any help! Thanks in advance

回答 (2)

fooks

2016-06-20 6:20 am

✔ 最佳答案

The weight of the block = mg = 2.2g N

Component of the weight of the block along the incline = 2.2g × sin25° N

Component of the weight of the block normally acting on the incline = 2.2g × cos25° N
Kinetic frictional force = 2.2g × cos25° × μ(k) N

When the block slides down the incline at a constant speed, resultant force acting on the block = 0
i.e. Kinetic frictional force = Component of the weight of the block normally acting on the incline
2.2g × cos25° × μ(k) = 2.2g × sin25°
μ(k) = sin25° / cos25°
μ(k) = tan25°
The coefficient of kinetic friction, μ(k) = 0.47

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[匿名]

2016-06-20 6:14 am

Yahoo is a wasteland

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