Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
In the Figure the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.3 kg...Asked by chirayu
In the Figure the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.3 kg, block B has a mass of 2.7 kg, and angle è is 26 °. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen 27 cm? and in ths figure block b is hanging down and block A is on inclined.
Answers
Answered by
drwls
The increased (and total) kinetic energy equals the lost potential emergy. If block B has fallen 0.27 m vertically, block A has moved up the incline for an altitude change of 0.27 sin 26 = 0.1184 m
Total P.E. lost = 2.7*9.8*0.27 - 1.3*9.8*0.1184 = 7.14 - 1.51 J = 5.63 J
That equals the total K.E. increase,
(1/2)(MA + MB)*V^2
Total P.E. lost = 2.7*9.8*0.27 - 1.3*9.8*0.1184 = 7.14 - 1.51 J = 5.63 J
That equals the total K.E. increase,
(1/2)(MA + MB)*V^2
Answered by
chirayu
thank u so much for help Drwls!!!!!!!!!!1
Answered by
Reading_Sucks
The increased (and total) kinetic energy equals the lost potential emergy. If block B has fallen 0.27 m vertically, block A has moved up the incline for an altitude change of 0.27 sin 26 = 0.1184 m
Total P.E. lost = 2.7*9.8*0.27 - 1.3*9.8*0.1184 = 7.14 - 1.51 J = 5.63 J
That equals the total K.E. increase,
(1/2)(MA + MB)*V^2
Total P.E. lost = 2.7*9.8*0.27 - 1.3*9.8*0.1184 = 7.14 - 1.51 J = 5.63 J
That equals the total K.E. increase,
(1/2)(MA + MB)*V^2
Answered by
Anonymous
jonga am dead
Answered by
As
In the figure, the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.0 kg, block B has a mass of 2.0 kg, and angle θ is 30°. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy in Joules when block B has fallen 7 cm? (take g = 10 m/s2).
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.