A mass m, = 6.5 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m, = 3.7 kg that hangs freely. 1) What is the magnitude of the acceleration of block 1? m/s Submit 2) What is the tension in the string? N Submit 3) m Now the table is tilted at an angle of 0 = 75° with respect to the vertical. Find the magnitude of the new acceleration of block 1. m/s Submit

Physics for Scientists and Engineers: Foundations and Connections
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Chapter5: Newton's Laws Of Motion
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m.
m2
A mass m, = 6.5 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m,
= 3.7 kg that hangs freely.
1) What is the magnitude of the acceleration of block 1?
m/s? [ Submit
+)
2) What is the tension in the string?
N Submit
3)
m,
m.
Now the table is tilted at an angle of 0 = 75° with respect to the vertical. Find the magnitude of the new acceleration of
block 1.
m/s Submit
+)
4) At what "critical" angle will the blocks NOT accelerate at all?
Submit
+)
..................... ....... .
................. ........ .......
Transcribed Image Text:m. m2 A mass m, = 6.5 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m, = 3.7 kg that hangs freely. 1) What is the magnitude of the acceleration of block 1? m/s? [ Submit +) 2) What is the tension in the string? N Submit 3) m, m. Now the table is tilted at an angle of 0 = 75° with respect to the vertical. Find the magnitude of the new acceleration of block 1. m/s Submit +) 4) At what "critical" angle will the blocks NOT accelerate at all? Submit +) ..................... ....... . ................. ........ .......
5) Now the angle is decreased past the "critical" angle so the system accelerates in the opposite direction. If 0 = 22°
find the magnitude of the acceleration.
m/s Submit
+)
6) Compare the tension in the string in each of the above cases on the incline:
Te at 75° = Tecritical = Te at 22°
Te at 75° > Tecritical
> Te at 22°
OTe at 75° < Teeritical Te at 22°
Submit
Transcribed Image Text:5) Now the angle is decreased past the "critical" angle so the system accelerates in the opposite direction. If 0 = 22° find the magnitude of the acceleration. m/s Submit +) 6) Compare the tension in the string in each of the above cases on the incline: Te at 75° = Tecritical = Te at 22° Te at 75° > Tecritical > Te at 22° OTe at 75° < Teeritical Te at 22° Submit
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