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The coefficient of friction between a box and an inclined plane is 0.3639. If the slope...

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mikerhyme | Student, Kindergarten | Salutatorian

Posted September 21, 2013 at 2:57 PM via web

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The coefficient of friction between a box and an inclined plane is 0.3639. If the slope of the plane with the horizontal is increased to 45 degrees, what is the acceleration of the box.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 21, 2013 at 3:04 PM (Answer #1)

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The coefficient of friction between a box and an inclined plane is 0.3639.

Let the mass of the box be M. The slope of the plane with the horizontal is equal to 45 degrees. There is a gravitational force on the box in the vertically downward direction. This is equal to M*g. The components of this force in a direction parallel to the plane and perpendicular to the plane are M*g*sin 45 and M*g*cos 45.

The frictional force on the box is M*g*cos 45*0.3639. The net force on the box is M*g*sin 45 - M*g*cos 45*0.3639. The resulting acceleration of the box is `9.8*(1/sqrt 2) - 9.8*(1/sqrt 2)*0.3639` = 4.407

The box accelerates down the inclined plane at 4.407  m./s^2

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