A 4.5-kg brick is suspended by a light string from a 2.0-kg pulley. The brick is released from rest and falls to the floor below. The pulley may be considered a solid disk of radius 1.5 m. What is the linear acceleration of the brick and the tension in the string?
2 answers
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james is wrong.
F=ma
mg-T=ma
(4.5)(9.8)-T=4.5a
Torque=I(alpha fish)
I=1/2MR^2
TR=1/2MR^2(a/R)
1.5T=(1/2)(2.0)(1.5^2)(a/1.5)
T=1a
substitute for T
44.1-a=4.5a
44.1=5.5a
a=8.02m/s^2
substitute for a
T=8.02N
F=ma
mg-T=ma
(4.5)(9.8)-T=4.5a
Torque=I(alpha fish)
I=1/2MR^2
TR=1/2MR^2(a/R)
1.5T=(1/2)(2.0)(1.5^2)(a/1.5)
T=1a
substitute for T
44.1-a=4.5a
44.1=5.5a
a=8.02m/s^2
substitute for a
T=8.02N