Asked by Jennifer
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?
Answers
Answered by
James
lol yoooo you in brahmia's physics class bruh pay attention in class yo
Answered by
rob
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
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