Three blocks are connected on the table as shown below. The coefficient of kinet

Question

Three blocks are connected on the table as shown below. The coefficient of kinetic friction between the block of mass m2 and the table is 0.430. The objects have masses of m1 = 5.00 kg,m2 = 1.35 kg, and m3 = 1.90 kg, and the pulleys are frictionless.

(b) Determine the acceleration of each object, including its direction.

(c) Determine the tensions in the two cords.

m1: magnitude _______ m/s2 direction up down m2: magnitude _______ m/s2 direction left right m3: magnitude _______ m/s2 direction up down

Explanation / Answer

friction force Ff = µmg = 0.430 * 1.3kg * 9.8m/s² = 5.478 N
net accelerating force Fa = (5kg - 1.9kg) * 9.8m/s² - 5.478 N = 24.902 N
total mass = 8.25 kg, so the system acceleration a = 24.902 N / 8.25 kg = 3.01 m/s²
All three masses accelerate at the same rate.
Obviously, 5 kg accelerates down, 1.35 kg to the left, and 1.9 kg up.
left T = 5kg * (9.8 - 3.01)m/s² = 33.95 N
right T = 1.9kg * (9.8 + 3.01)m/s² = 24.339 N

Check: net force on 1 kg mass is
(33.95 - 24.339 - 5.478)N = 4.133 N
and therefore acceleration a = 4.133 m/s²

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