Question
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
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
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
Related Questions
"A 2.5-kg block slides down a 25 degree inclined plane with constant acceleration. The block start...
A 2.5-kg block slides down a 17.0° inclined plane with constant acceleration. The block starts from...
An 8kg block starts from rest from the top of a plane, inclined at 40o with respect to the horizonta...