A m1=5.0 kg block is suspended from a light string that passes over a pulley whose other end is 12 pts attached to m2=4.0 kg resting on a table with negligible friction.

A m1=5.0 kg block is suspended from a light string that passes over a pulley whose other end is 12 pts attached to m2=4.0 kg resting on a table with negligible friction. The pulley is a solid disk having radius R=0.20 m and mass MP=3.0 kg. The system is released from rest

a) What is the acceleration of mass m2 as it moves?

b) What are the tensions in the strings?

c) Through how many revolutions does the pulley turn if m2 slides for 0.78 M, seconds?

Solutions

Expert Solution

m₂a = m₂g - T...........(1)

T = m₂ (g-a)

T = m₁a............(2)

m₂(g-a) = m₁a

m₂g-m₂a=m₁a

a = 4.35 m/s²

Tensions in the string

T = m₁* a

T = 5 * 4.35

T1= 21.50 N

T₂ = m₂ (g - a )

T₂ = 4 * (9.8-4.35)

T₂ = 21.80 N

T=0.78 sec

2*3.14*0,20/0.78

v =1.6 m/s

revolutions per minute = speed in meters per minute / circumference in meters.

Circumference=2r

genius_generous answered 1 year ago

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