Question

Two blocks of mass 4kg and 6kg are attached by a spring constant k=200N/m both blocks are moving with same acceleration find elongation in spring ​

Answer 1

Given:

Mass of two blocks = 4 kg and 6 kg.

Spring constant k = 200 N/m

To find:

Elongation in the spring if the two blocks are moving with same acceleration.

Solution:

Let the elongation in the spring = 6 kg

And the acceleration be a m/s^2

For 6 kg block, as observed from the FBD the equation will be:

20 - kx = 6a     (1)

For the 4 kg block, the equation will be:

kx = 4a    (2)

On adding both the equations we get:

20 = 10a

a = 2 m/s^2

From equation 2:

kx = 4a

4*2 = 200* x

x = 0.04 m

Therefore the elongation in the spring will be of 2 m.