As shown in the figure, a block of mass M1 = 0.490 kg is initially at rest on a
ID: 2150213 • Letter: A
Question
As shown in the figure, a block of mass M1 = 0.490 kg is initially at rest on a slab of mass M2 = 0.840 kg, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass M3. The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction ?k = 0.380 and a coefficient of static friction ?s = 0.520 with both the table and the block. When released, M3 pulls on the string and accelerates the slab, which accelerates the block. Find the maximum mass of M3 that allows the block to accelerate with the slab, without sliding on top of the slab.
As shown in the figure, a block of mass M1 = 0.490 kg is initially at rest on a slab of mass M2 = 0.840 kg, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass M3. The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction ?k = 0.380 and a coefficient of static friction ?s = 0.520 with both the table and the block. When released, M3 pulls on the string and accelerates the slab, which accelerates the block. Find the maximum mass of M3 that allows the block to accelerate with the slab, without sliding on top of the slab.Explanation / Answer
m3g -T = m3a ....(1) T- 0.38*(m1+m2)g = (m1+m2)a......(2) 0.52*m1g = m1a....(3) condition for not sliding on slab m2 solving 1 and 2 m3= [(0.38*(m1+m2)g+(m1+m2)a)/(g-a)]....(4) from 3 ==> a = 0.52g = 5.096 plug a=5.096 in 4 ==> m3 = 2.49 kg
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