Question

The block shown in figure (5.11) has a mass M and
descends with an acceleration a. The mass of the string
below the point A is m. Find the tension of the string at
the point A and at the lower end​

Answer 1

Explanation:

Let the tension in the string going over pulley be T

And the acceleration of mass m is a (upward direction)

For left side mass,

T−mg=ma

Or ,T=mg+ma

For right side masses,

5mg−T=5ma

Or ,T=5mg−5ma

From above equations,

mg+ma=5mg−5ma

Or ,6ma=4mg

Or ,a=2g/3

hope it helps you

Answer 2

Answer:

a=2g/3

is the answers.....