A mass m moving with velocity v collides a spring mass system at rest on a smooth floor as shown in the figure. The maximum force developed in the spring

The question seems to be incomplete. However, I can provide a general approach to solve such problems.

1. First, we need to understand that when the mass m collides with the spring, it will compress the spring. The maximum force in the spring will be when the spring is at its maximum compression.

2. The energy of the mass m before the collision is kinetic energy, which is given by (1/2)mv^2.

3. When the spring is at its maximum compression, all the kinetic energy of the mass m will have been transferred into potential energy in the spring. The potential energy stored in a spring is given by (1/2)kx^2, where k is the spring constant and x is the displacement (or compression in this case).

4. By conservation of energy, we can equate the kinetic energy before the collision to the potential energy at maximum compression. This gives us (1/2)mv^2 = (1/2)kx^2.

5. We can solve this equation for x, the maximum compression of the spring: x = sqrt((mv^2)/k).

6. The force developed in the spring is given by Hooke's law, F = kx. Substituting our expression for x gives F = ksqrt((mv^2)/k) = sqrt(kmv^2).

So, the maximum force developed in the spring is sqrt(k*mv^2). This is under the assumption that the collision is perfectly elastic and there are no other forces acting on the system. If there are other forces or energy losses, those would need to be taken into account.