"two bars connected by a wtless spring of stiffness x and length (in the non deformed state) L rest on a horizonntal plane. A constant force F starts acting on one of the bars. Find the maximum and minimum distances between the bars during the subsequent motion of the system, if the masses of the bars are equal"

 

explain why will the spring stretch on a frictionles plane.

If mass of the bars be m₁ and m₂. Mass m₁ be at the left side and m₂ be attached to right side of the spring.

 

force applied on the system = F on m₁ in the direction left to the spring.

 

So, acceleration in the system = a = F/(m₁+m₂)

 

This frame is accelerated and thus non inertial frame of reference, and in order to apply Newton's laws of motion we apply a pseudo force (m₁a) towards left on the block m₁ and m₂a towards left to the block m₂.

 

We consider the problem from Center of mass frame. In this frame bars move in opposite directions and come to instantaneous rest at some instant. The elongation of the spring will be minimum and maximum at this instant.

 

Let bar m₁ is displaced by a distance x₁ and the bar m₂ by x₂ from the initial positions.

 

Applying energy equation in center of mass frame,

 

T + U = W_ext

 

Here, Wext = work one by the pseudo forces

 

U = Potential energy of the spring = k(x₁ +x₂)²/2

 

T = Kinetic energy = 0

 

W_ext = m₁F(x₁+x₂)/(m₁+m₂),

 

So, k(x₁ +x₂)²/2 = m₁F(x₁+x₂)/(m₁+m₂)

 

or, (x₁+x₂) = 0 or (x₁+x₂) = 2m₁F/k(m₁+m₂)

 

Hence the maximum seperation b/w the bars equals : L + 2m₁F/k(m₁+m₂)

 

The minimum separation corresponds to zero elongation and is equal to L.

 

 

This can easily be explained as there is accleration in the mass bar m₂ as a result of which there is tension in the spring.