Asked by Daniel
A mass of 10kg is placed on the top of a plane inclined at 45 degrees to the horizontal and allowed to slide down. What is the acceleration of the mass if the frictional force opposing is 3.2N.
Answers
Answered by
Henry
mg = 10kg * 9.8 = 98N @ 45deg.
Fp = 98*sin(45) = 69.3N = Force acting parallel to plane and downward.
Fn = Fp - Ff = 69.3 - 3.2 = 66.1N.
a = Fn / m = 66.1 / 10kg = 6.61m/s^2.
Fp = 98*sin(45) = 69.3N = Force acting parallel to plane and downward.
Fn = Fp - Ff = 69.3 - 3.2 = 66.1N.
a = Fn / m = 66.1 / 10kg = 6.61m/s^2.
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