For the apparatus shown in the figure, there is no slipping between the cord and

Question

For the apparatus shown in the figure, there is no slipping between the cord and the surface of the pulley. The blocks have mass of 3.0 kg and 5.7 kg, and the pulley has a radius of and a mass of 10.3 kg. (i) What is the acceleration of the two blocks? (ii) What is the tension on both sides of the cord? (iii) At the instant the 5.7 kg mass has fallen 1.5 m starting from rest, find the speed of each block. (Assume the pulley is in the shape of a uniform solid disk and has no friction in its axle. The moment of inertia of a disc is MR2/2.)

Explanation / Answer

step;1

Given that

mass m1=3 kg

mass m2=5.7 kg

mass of the pulley M=10.3 kg

step;2

now we find the acceleration of the two blocks

the acceleration a=(m2-m1)g/(m1+m2+M/2)

=(5.7-3)*9.8/(3+5.7+10.3/2)

=26.46/13.85

=1.9 m/s^2

now we find the tension on both sides of the cord

the tension T1=m1(g+a)=3(9.8+1.9)=35.1 N

the tension T2=m2(g-a)=5.7(9.8-1.9)=45.03 N

step;3

now we find the velocity of the block

according to kinematices equation

v^2=u^2+2as

v^2=0+2*1.9*1.5

v^2=5.7

v=2.4 m/s

the velocity of each block v=2.4 m/s

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