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
2 answers
Calculate the coefficient of sliding friction between the crate and the floor
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.
1 answer