2 blocks m1 (10kg) and m2 (30kg) are linked by a rope and are being pulled across?

Question

a surface via the 89N tension force in the tow rope (which is pulling horizontally). The two blocks are made of different substances; each has its own kinetic friction, uk with the surface. uk of mass1 is 0.2 uk of mass2 is 0.1 The tension in the rope linking the two blocks is: 1) 10N
2) 20N
3)22.3 N
4)29.6 N
5)36.9 N

Answer 1

Draw a free body diagram as shown below:

[m1]------> T<---[m2]------->89N
<-------- <---------
f1 =19.6N f2=29.4N


First solve for f1 and f2:

f=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values

f1=0.2*10*9.8
=19.6N to the left
f2=0.1*30*9.8
=29.4N to the left

Looking at the free body diagram:

for m1:

net force, f1=T-19.6

From the formula, f=ma where f is the net force, m the mass, and a is the acceleration.:

f1=10a

Thus,

T-19.6=10a
a=(T-19.6)/10

for m2:

Net force, f2=89-29.4-T
=59.6-T

Similarly as for m1:

59.6-T=30a
a=(59.6-T)/30

Now a for m1 is also the same a as for m2, therefore:

(T-19.6)10=(59.6-T)/30
The zero on each side cancels out.
3T-58.8=59.6-T
4T=59.6+58.8
4T=118.4
T=29.6N in the direction shown in the above free body diagram.