Fn = mg*cos45 = 15 * 9.8 * cos45=104 N.
= Normal force = Force perpendicular to
the incline.
Fk = u*Fn = 0.2 * 104 = 20.8 N.
A 15.0-kg box is released from rest on a rough inclined plane tilted at an angle of 45° to the horizontal. The coefficient of kinetic friction between the box and the inclined plane is 0.200. What is the force of kinetic friction on the box?
Answer
Fifteen square root of two Newtons
Thirty square root of two Newtons.
Fifteen square root of three over two Newtons.
Fifteen square root of three Newtons.
Thirty square root of three Newtons.
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