Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.

Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.

Block 1 is removed without disturbing block 2.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is mass of block 3

Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a

Making T the subject gives us;

T = (m2)g - (m2)a - - - (eq 2)

where;

g is acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is mass of block 2

To solve for the acceleration, we will just substitute (m3)a for T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)