A block of mass M and a block of mass m are connected by a cord that passes over a frictionless pulley. This is known as an Atwood’s machine. The heavier block falls to the ground. What is the acceleration of the blocks?

The free-body diagrams of the two blocks are shown above. T is the tension in the cord.

We can write the following equations:

Ma = Mg − T

ma = T − mg

Adding these two equations together,

Ma + ma = Mg − mg

(M + m)a = (M − m)g

a = (M − m)g / (M + m)

The gravitational potential energy of a body that has dropped the distance y is:

GPE = GPEi − mgy

GPEi is the initial gravitational potential energy of the body.

In the above figure, the heavier block has dropped the distance y and the lighter block has risen the distance y.

The total gravitational potential energy of the two blocks is:

GPE = GPEi + mgy − Mgy

The total kinetic energy of the two blocks is:

KE = ½ Mv ² + ½ mv ²

The derivative of gravitational potential energy is:

DGPE = mg − Mg

The derivative of kinetic energy is:

DKE = Ma + ma

The total energy of the two blocks remains the same as the blocks move. The rate of change of energy is zero. We have:

DGPE + DKE = 0

mg − Mg + Ma + ma = 0

Ma + ma = Mg − mg

(M + m)a = (M − m)g

a = (M − m)g / (M + m)