Question

Two blocks of masses 4kg and 2kg are attached to ends of a light spring having spring constant 1200 N/m. The whole system is placed on the smooth horizontal surface. When the spring is in relaxed state, the 4kg block is given velocity as shown below...”?”

The maximum extension in the spring is?

1, 10 cm
2. 12 cm
3. 20 cm
4. `sqrt(10)` cm

Answer 1

As,the `4 Kg` block is provided with a velocity,it will keep on compressing the spring,and the restoring force provided by the spring will try to push the block away from each other.

As,here restoring force by the spring acts as `F=-kx` (i.e against the direction of motion of the blocks)

We can consider the system to perform S.H.M

So,this `3 m/s` velocity will be the maximum velocity of this S.H.M ,as on going further towards the spring,the velocity will decrease due to resistive force by the spring.

Now, here reduced mass of the system(`m`) is `4*2/(4+2) Kg = 8/6 Kg`

So, `omega = sqrt(K/m) = sqrt(1200*6/8)`

Now,maximum velocity of S.H.M is expressed as `V=Aomega` (where, `A` is the amplitude of motion,and here in this question maximum extension of the spring)

Putting ,`V= 3` we get,

`9 = A^2 * 900`

or, `A= 1/10 m`

so, maximum extension will be `1/10 *100 cm =10 cm`