#$&*
Phy 121
Your 'cq_1_12.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Masses of 5 kg and 6 kg are suspended from opposite sides of a light frictionless
pulley and are released.
What will be the net force on the 2-mass system and what will be the magnitude and
direction of its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet=5kg*9.8m/s^2=49N
Fnet=6kg*9.8m/s^2=58.8N
a=107.8N/11kg=9.8m/s^2
#$&*
@& There is only one net force on the system.
There are forces of 49 N and 58.8 N, but they pull the system in two different directions. If you combine these two forces, taking account of the fact that they pull in opposite directions, you get the net force.*@
If you give the system a push so that at the instant of release the 5 kg object is
descending at 1.8 meters / second, what will be the speed and direction of motion of
the 5 kg mass 1 second later?
answer/question/discussion: ->->->->->->->->->->->-> :
-9.8m/s^2
#$&*
@& To get the acceleration you have to divide the net force by the total mass of the system. This will then change your answer to this question.*@
*#&!
@& See my notes and spend up to 10 minutes giving this another try. Whatever you get in that time, send it and I'll follow up.*@
#$&*
Phy 121
Your 'cq_1_12.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Masses of 5 kg and 6 kg are suspended from opposite sides of a light frictionless
pulley and are released.
What will be the net force on the 2-mass system and what will be the magnitude and
direction of its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet=5kg*9.8m/s^2=49N
Fnet=6kg*9.8m/s^2=58.8N
a=107.8N/11kg=9.8m/s^2
#$&*
@& There is only one net force on the system.
There are forces of 49 N and 58.8 N, but they pull the system in two different directions. If you combine these two forces, taking account of the fact that they pull in opposite directions, you get the net force.*@
If you give the system a push so that at the instant of release the 5 kg object is
descending at 1.8 meters / second, what will be the speed and direction of motion of
the 5 kg mass 1 second later?
answer/question/discussion: ->->->->->->->->->->->-> :
-9.8m/s^2
#$&*
@& To get the acceleration you have to divide the net force by the total mass of the system. This will then change your answer to this question.*@
*#&!
@& See my notes and spend up to 10 minutes giving this another try. Whatever you get in that time, send it and I'll follow up.*@