Okay I\'m down to my last two questions for this homework, and I have no idea ho

Question

Okay I'm down to my last two questions for this homework, and I have no idea how to figure them out. Any help would be greatly appreciated. Thank you!

Objects with masses m1 = 11.0 kg and m2 = 6.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m2 falls 1.00 m in 1.57 s, determine the coefficient of kinetic friction between m1 and the table.

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Objects with masses m1 = 11.0 kg and m2 = 6.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m2 falls 1.00 m in 1.57 s, determine the coefficient of kinetic friction between m1 and the table. The coefficient of static friction between the m = 2.85-kg crate and the 35.0½ incline of the figure below is 0.350. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Explanation / Answer

Number 1)

First we need the acceleration

d = vot + .5at2

1 = 0 + .5(a)(1.57)2

a = .811 m/s2

Then the sum of forces on m2...

mg - T = ma

6(9.8) - T = 6(.811)

T = 53.9 N

For the sum of forces on m1...

T - uFn = ma

53.9 - u(11)(9.8) = 11(.811)

u = .417

Number 2)

The sum of forces on the mass...

The force applied and friction will acts up the plane. A component of the weight will act down the plane, so...

F + umgcos(angle) = mgsin(angle)

F + (.35)(2.85)(9.8)(cos 35) = (2.85)(9.8)(sin 35)

F = 8.01 N

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