Blocks A and B are connected by a taut string that passes over a frictionless pu

Question

Blocks A and B are connected by a taut string that passes over a frictionless pulley. The blocks are initially moving at the same speed such that B goes down while A goes up the inclined plane. Each mass is 30Kg. and the coefficient of kinetic friction between block A and the incline is 0.4 The incline angle is 30 degrees. a) Using the work/energy principle and motion formulas, calculate the magnitude and direction of the acceleration of each block. b) Also, calculate the tension in the string. 4) Same setup as number 3, but the mass of A is 40Kg and the mass of B is 10Kg.

Explanation / Answer

mass, m = 30 kg
friction coefficient, mu = 0.4
theta = 30 deg
when the block B comes down by distance s, the block A slides up by distance s
so, work done by gravity on A = -mgs*sin(theta)
work done by gravity on B = mgs
work done by friction on A = fs
but f = mu*mgcos(theta)
so net work done on the system = change in KE ( work energy theorem)
let initial velocity of the systme be u and the final velocity be v
then
-mgs*sin(theta) + mgs - mu*mg*s*cos(theta) = 0.5*2*m[v^2 - u^2]
now, from kinematics equations
2*a*s = v^2 - u^2
so,
-mgs*sin(theta) + mgs - mu*mg*s*cos(theta) = 2a*m*s
-g*sin(theta) + g - mu*g*cos(theta) = 2a = g(1 - sin(theta) - mu*cos(theta) ) = 1.5067 m/s/s
a = 0.75335 m/s/s

now let tension in the string be T
then from blovk B
mg - T = ma
T = m(g-a) = 30(9.81 - 1.5067/2) = 271.6995

for the next setup
Mb = 10 kg
Ma = 40 kg

so, work done by gravity on A = -Mags*sin(theta)
work done by gravity on B = Mbgs
work done by friction on A = fs
but f = mu*Ma*gcos(theta)
so net work done on the system = change in KE ( work energy theorem)
let initial velocity of the systme be u and the final velocity be v
then
-mMa*gs*sin(theta) + Mb*gs - mu*Ma*g*s*cos(theta) = 0.5(mA + mB)[v^2 - u^2]
now, from kinematics equations
2*a*s = v^2 - u^2
so,
-Ma*gsin(theta) + Mb*g - mu*Ma*g*cos(theta) = (mA + mB)a
-40*gsin(30) + 10*g - 0.4*40*g*cos(30) = (50)a
a = -11.346 m/s/s

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