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A block slides with a constant velocity down a plane inclined at 37 to the horizontal....

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jul17 | Student | eNoter

Posted November 13, 2011 at 8:14 PM via web

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A block slides with a constant velocity down a plane inclined at 37 to the horizontal. What is the coefficient of kinetic friction between the block and the plane?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 14, 2011 at 1:37 AM (Answer #1)

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The gravitational force of attraction acting on the block pulls it down with a force equal to F = m*g, where g = 9.8 m/s^2 and m is the mass of the block.

The block slides down the plane inclined at 37 degrees to the horizontal at a constant velocity. The acceleration in this case is 0.

The force acting downwards can be split into two components, one acting along the the inclined plane and the other perpendicular to the plane. The component along the plane is F*sin 37. The component perpendicular to the plane is F*cos 37. If the coefficient of kinetic friction is k, kinetic friction is N*k, where N is the normal force. Here N = F*cos 37. Equating the two forces:

F*cos 37*k = F*sin 37

=> k = sin 37/cos 37

=> k = tan 37

=> k = 0.7535

The required coefficient of kinetic friction is 0.7535

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