Asked by Okiki
A crate of mass 42kg moving with a speed of 8m-1 on a rough horizontal floor is brought to rest after sliding a distance of 5m on the floor.calculate the coefficient of sliding friction between the crate and the floor
Answers
Answered by
Okiki
Calculate the coefficient of sliding friction between the crate and the floor
Answered by
henry2,
F = m*g = 42 * 9.8 = 411.6 N. = force of crate = normal force(Fn).
Fp = F*sin A = 411.6*sin 0 = 0 = Force parallel with plane.
V^2 = Vo^2 + 2a*d = 0.
8^2 + 2a*5 = 0,
a = -6.4 m/s^2.
Fp - u*Fn = M*a.
0 - u*411.6 = 42 * (-6.4),
u = 0.653. = Coefficient of friction.
Fp = F*sin A = 411.6*sin 0 = 0 = Force parallel with plane.
V^2 = Vo^2 + 2a*d = 0.
8^2 + 2a*5 = 0,
a = -6.4 m/s^2.
Fp - u*Fn = M*a.
0 - u*411.6 = 42 * (-6.4),
u = 0.653. = Coefficient of friction.
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