TBE physics chapter 2 passage 1 question 5

[2010premed]

If the force of kinetic friction acting on M1 goes to zero, the resulting NET forces on M1 and M2 will be:

A. less than M1g and less than M2g
B. less than M1g and more than M2g
C. more than M1g and less than M2g
D. more than M1g and more than M2g

The problem is a pulley with M1 resting on a block and is connected to a rope which is connected to a pulley and M2 is hanging in the air with the same rope from the pulley. Thanks!

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[BerkReviewTeach]

If the force of kinetic friction acting on M1 goes to zero, the resulting NET forces on M1 and M2 will be:

A. less than M1g and less than M2g
B. less than M1g and more than M2g
C. more than M1g and less than M2g
D. more than M1g and more than M2g

The problem is a pulley with M1 resting on a block and is connected to a rope which is connected to a pulley and M2 is hanging in the air with the same rope from the pulley. Thanks!

If you first consider the hanging mass, M2, it has forces of M2g pulling it down and tension pulling it up. No matter what, because tension opposes the force of gravity, the net force acting on M2 must be less than M2g, which cancels out choices B and D.

The two masses are connected by a rope, so they experience the same magnitude of a, meaning a1 = a2. a2 is less than g, so 1 must also be less than g. This means that the force causing M1 to accelerate is M1a, which must be less than M1g. Only choice A works.

 

[2010premed]

Thanks!