m1 = 2.7 kg block slides on a frictionless horizontal surface and is connected on one side to a spring (k = 40 N/m) as shown in the figure above. The other side is connected to the block m2 = 3.6 kg that hangs vertically. The system starts from rest with the spring unextended.

a) What is the maximum extension of the spring?

To find the maximum extension of the spring, we need to consider the conservation of energy. As the system starts from rest, the initial kinetic energy is zero. The only form of energy in the system is potential energy.

Initially, the spring is unextended, so its potential energy is zero. As the block m2 falls, it gains gravitational potential energy, which is given by m2gh, where m2 is the mass of the hanging block, g is the acceleration due to gravity, and h is the height through which the block m2 falls.

At the same time, the spring gets compressed by a certain distance x, which stores potential energy. The potential energy stored in the spring is given by (1/2)kx^2, where k is the spring constant and x is the distance by which the spring is compressed.

At the maximum extension, all of the gravitational potential energy of m2 is converted into potential energy stored in the spring. Therefore, we can equate the two expressions for potential energy:

m2gh = (1/2)kx^2

Substituting the given values:

m2 = 3.6 kg
g = 9.8 m/s^2
k = 40 N/m

Replacing these values in the equation and solving for x, we can determine the maximum extension of the spring.