A 4 kg block lies on a horizontal table whose coefficient of friction is 0.3. A

Question

A 4 kg block lies on a horizontal table whose coefficient of friction is 0.3. A horizontal rope is redirected by a uniform solid disk (pulley) of mass 2 kg to a mass of 12 kg hanging off the table. The radius of the solid disk is 10 cm and the rope doesn't slip on the disk.
a) What is the linear acceleration of the 4 kg block?
b) What is the angular acceleration of the disk?
c) What is the tension in the rope?
d) how far does the top block travel in 2 s?
e) How many revolutions does the disk of the pulley make in 2 s?
f) How far does a point on the edge of the disk of the pulley travel in 2 s?

I know this is a REALLY long one, and I really appreciate any help with it! I am totally drowning in this class and don't have time to meet with my tutor before this assignment is due :/
THANK YOU!!

A 4 kg block lies on a horizontal table whose coefficient of friction is 0.3. A horizontal rope is redirected by a uniform solid disk (pulley) of mass 2 kg to a mass of 12 kg hanging off the table. The radius of the solid disk is 10 cm and the rope doesn't slip on the disk. What is the linear acceleration of the 4 kg block? What is the angular acceleration of the disk? What is the tension in the rope? how far does the top block travel in 2 s? How many revolutions does the disk of the pulley make in 2 s? How far does a point on the edge of the disk of the pulley travel in 2 s?

Explanation / Answer

inertia of pulley = (0.5)*m*r^2 = 0.5*2*0.1^2 = 0.01 kg m^2

let the linear accelearion be a ..

so... angular acceleration = a / r = a / 0.1 = 10 a

let the tension in rope connecting 4 kg block be T1 and that in 12 kg block be T2

so... (T2 - T1)*radius = inertia * angular acc

so.. ( T2 - T1 )*0.1 = 0.01 * 10 a

so... T2 - T1 = a

for 4 kg block..

T1 - (0.3*4*9.8) = 4*a

so.. T1 = 4a + 11.76

so... T2 = T1 + a = 5a + 11.76

for 12 kg block..

12*9.8 - T2 = 12*a

12*9.8 - 5a - 11.76= 12a

so.. a = ((12*9.8)-11.76)/17 = 6.22588 m/sec2

b) angular accelelraion of disk = 10*a = 62.2588 rad/sec2

c) Tension T1 = 4a = 4*6.22588 = 24.90353 N

Tension T2 = 5a = 5*6.22588 = 31.1294 N

d) for top block

a = 6.22588

u = initial vel = 0

t = 2 sec

so distance s = ut + 0.5*a*t^2 = 0 + (0.5*6.22588*2^2) =12.45176 m

e) revolution = s / (2*pi*rad) = 12.45176 / (2*pi*0.1) = 19.81759 revs

f) distance travelle by a point on the edge of the disc = s = 12.45176 m

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