Question

The block has uniform mass density. The tension at mid-
point of the block is (M = 6 kg)
a 8N
b 14N
c 12N
d 20N​

Answer 1

Answer:

We begin by examining a slightly different problem: similar to this figure but without the string. The motivation is that if (without the string) block A is found to accelerate faster (or exactly as fast) as block B then (returning to the original problem) the tension in the string is trivially zero. In the absence of the string,

a

A

=F

A

/m

A

=3.0m/s2

a

B

=F

B

/m

B

=4.0m/s

2

so the trivial case does not occur. We now (with the string) consider the net force on the system: Ma=F

A

+F

B

=36N. Since M=10kg (the total mass of the system) we obtain a=3.6m/s

2

. The two forces on block A are F

A

and T (in the same direction), so we have

m

A

a=F

A

+T⇒T=2.4N.