Two blocks are arranged at the ends of a
massless string as shown in the ﬁgure. The
system starts from rest. When the 4.25 kg
mass has fallen through 0.33 m, its downward
speed is 1.25 m/s.
The acceleration of gravity is 9.8 m/s
2
.What is the frictional force between the
5.76 kg mass and the table?
Answer in units of N

Question

Two blocks are arranged at the ends of a
massless string as shown in the ﬁgure. The
system starts from rest. When the 4.25 kg
mass has fallen through 0.33 m, its downward
speed is 1.25 m/s.
The acceleration of gravity is 9.8 m/s
2
.What is the frictional force between the
5.76 kg mass and the table?
Answer in units of N

anonymous · Answer

where is the figure??

anonymous · Answer

attachment

anonymous · Answer

42.8 N 
i'm not sure

anonymous · Answer

i think first we need to calculate net acceleration of the system.
In fact question is easy, but i don't remember how to do. It has been a very long time solving these type of question

anonymous · Answer

I don't have the screen shot, but I imagine one block m1 is on the table, and the other block, m2 is falling. The acceleration of the pair is determined by the net force, that of gravity on m2 minus the frictional force on m1. The net force provides F = (m1 + m2) a, and we can calculated a from the equation for acceleration under constant forces, here v^2=2 a s. I think that will do it.

anonymous · Answer

@douglaswinslowcooper i think u are right.