Ablock with mass m1 = 9.4 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of μk = 0.52. a second block with a mass m2 = 11.5 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. the second block is released and motion occurs.

a) write an expression for the magnitude of acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (assume the cable connecting the masses is ideal)
b) write an expression using the variable provided for the magnitude of the tension force, t.
c) what is the tension, t, in newtons?

a)

Let's write the equation of the forces for the two blocks.

For block m1, the one which is on the horizontal table, there are two forces: the tension of the string, T, and the force of friction, , pulling in the opposite direction. So the equation of motion is

(1)

For block m2, the one which is hanging vertically, there are two forces: the weight , pushing downward, and the tension in the cable, acting upward. So the equation of the forces is

(2)

(note that if the cable is inextensible, the two blocks have same acceleration, so ).

We can isolate the tension T from eq.(1):

And substitute it into eq. (2):

b)

Since the accelerations of the two blocks are equal: , we can rewrite the two equations of motion as

(1)

(2)

So we can substitute the expression for a from (2) into (1) and solve the equation for T, the tension:

c) 77.0 N

In order to find the magnitude of the tension, we just need to substitute data into the equation find at point b):