A 4 kg block slides down a rough inclined plane inclined at 30° with the horizontal. Determine the coefficient of kinetic friction between the block and the surface if the block has an acceleration of 1.2 m/s2.
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M*g = 4 * 9.8 = 39.2 N.
Fp = 39.2*sin30 = 19.6 N. = Force parallel to the incline.
Fn = 39.2*Cos30 = 33.9 N. = Normal force.
Fk = u*Fn = u*33.9 = 33.9u.
Fp-Fk = M*a.
19.6 - 33.9u = 4 * 1.2
-33.9u = 4.8-19.6 = -14.8
u = 0.437.
Fp = 39.2*sin30 = 19.6 N. = Force parallel to the incline.
Fn = 39.2*Cos30 = 33.9 N. = Normal force.
Fk = u*Fn = u*33.9 = 33.9u.
Fp-Fk = M*a.
19.6 - 33.9u = 4 * 1.2
-33.9u = 4.8-19.6 = -14.8
u = 0.437.
2 answers