Asked by Khadija
A 2.6 kg block slides down a 25° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity is 0.75 m/s. The incline is 1.6 m long.
(a) What is the acceleration of the block?
(b) What is the coefficient of friction?
(a) What is the acceleration of the block?
(b) What is the coefficient of friction?
Answers
Answered by
Henry
a. Wb = M*g = 2.6kg * 9.8N/kg=25.48 N. =
Weight of block.
Fp = 25.48*sin25 = 10.77 N. = Force
parallel to the inclined plane.
Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the
incline.
a=V^2-Vo^2)/2d
a = (0.75^2-0)/2.6 = 0.2163 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6 * 0.2163 = 0.5625
Fk = 10.77 - 0.5625 = 10.21 N. = Force
of kinetic friction.
u = Fk/Fn = 10.21/23.09 = 0.4421
Weight of block.
Fp = 25.48*sin25 = 10.77 N. = Force
parallel to the inclined plane.
Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the
incline.
a=V^2-Vo^2)/2d
a = (0.75^2-0)/2.6 = 0.2163 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6 * 0.2163 = 0.5625
Fk = 10.77 - 0.5625 = 10.21 N. = Force
of kinetic friction.
u = Fk/Fn = 10.21/23.09 = 0.4421
Answered by
Henry
Correction:
a = (0.75^2-0)/3.2 = 0.1758 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6*0.1758 = 0.4570
Fk = 10.77 - 0.4570 = 10.31 N
u = Fk/Fn = 10.31/23.09 = 0.4466
a = (0.75^2-0)/3.2 = 0.1758 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6*0.1758 = 0.4570
Fk = 10.77 - 0.4570 = 10.31 N
u = Fk/Fn = 10.31/23.09 = 0.4466
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