Question

The block of mass m₁ shown in figure (12-E2) is fastened to the spring and the block of mass m₂ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k)(m₁+m₂)g.sinθ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of the blocks at the time of separation?

Answer 1

(a) Due to the gravitational force on the blocks that's will equal the spring force

So,
kx = (m1+m2)g sin theta

So,
x = (m1+m2)gsin theta/k

Now,
(b) The blocks are further pushed 2/k(m1+m2)g sin theta
So, Total compression = 3/k(m1+m2)g sin theta
That means that The amplitude of the SHM followed will be 2/k(m1+m2)g sin theta

And, after the total compression in the spring will become zero that is the spring comes to it's natural length , then the blocks will leave each other's contact.

(c) By work energy theorem,

Change in Kinetic Energy = Work done by all forces,

So,
1/2*(m1+m2)*v^2 = 1/2*k*x^2 - (m1+m2)gsin theta*x
Therefore,
1/2*(m1+m2)*v^2 = 1/2*k*(3/k(m1+m2)g sin theta)^2 - (m1+m2)g sin theta *(3/k (m1+m2) g sin theta)
So,
v = g sin theta * √(3/k(m1+m2)

Hope this helps you !

Answer 2

Answer is in attachment

thank you