023 (part 1 of 2) 10.0 points A block of mass 1 kg and one of mass 8 kg are conn
ID: 2039874 • Letter: 0
Question
023 (part 1 of 2) 10.0 points A block of mass 1 kg and one of mass 8 kg are connected by a massless string over a pulley that is in the shape of a disk having a radius of 0.33 m, and a mass of 4 kg. In addition, the blocks are allowed to move on a fixed block- wedge of angle 44°, as shown. The coefficient of kinetic friction is 0.27 for both blocks A dis mo co res 0.33 m 4 kg 1 kg Ho ac 44° What is the acceleration of the two blocks? The acceleration of gravity is 9.8 m/s2. As- sume the positive direction is to the right. Answer in units of m/s2. pa 024 (part 2 of 2) 10.0 points Find the tension in the horizontal part of the string. Answer in units of N. friExplanation / Answer
a)
let m1 = 1kg, m2 = 8 kg
M = 4 kg, r = 0.33 m
let T1 is the tension in the string connected to m1, T2 is the tension in the string connected to m2 and a is the acceleration of the blocks.
net force acting on m1, Fnet1 = T1 - fk
m1*a = T1 - mue_k*m1*g
T1 = m1*a + mue_k*m1*g
T1 = 1*a + 0.27*1*9.8
T1 = a + 2.646 -----------(1)
Net force acting on m2, Fnet2 = m2*g*sin(44) - T2 - fk
m2*a = m2*g*sin(44) - T2 - mue_k*N2
m2*a = m2*g*sin(44) - T2 - mue_k*m2*g*cos(44)
T2 = m2*g*sin(44) - mue_k*m2*g*cos(44) - m2*a
= 8*9.8*sin(44) - 0.27*8*9.8*cos(44) - 8*a
T2 = 54.46 - 15.23 - 8*a ----(2)
Net torque acting on the pulley, Tnet = I*alfa
(T2 - T1)*r = I*a/r
(T2 - T1)*r = 0.5*M*r^2*a/r
T2 - T1 = 0.5*M*a
54.46 - 15.23 - 8*a - (a + 2.646) = 0.5*M*a
54.46 - 15.23 - 8*a - (a + 2.646) = 0.5*4*a
a = 3.32 m/s^2 <<<<<<<<<<<-----------------Answer
b) from equation 1
T1 = 1*a + 2.646
= 1*3.32 + 2.646
= 5.97 N <<<<<<<<<<<-----------------Answer
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