Question

Q.21
In figure shown, pulleys are ideal m. > 2 m. Initial
the system is in equilibrium and string connecting
to rigid support below is cut. Find the init
acceleration of m?
ele
m.​

Answer 1

Let the extension in spring be x

Then m1 moves down by 2x .  

By energy conservation m1gx=12kx2

Hence x= (4mg)/(k)

So PE is 8m2g2k

But the answer given as 2m2g2k

If m2 is attached to the ground then it doesn't really matter what value it has, provided that it is less than m1. The string or wire which attaches m2 to the ground will supply enough force to counteract the weight of m1. (If m2>m1 then there is no equilibrium.)

At equilibrium the tension in the string holding up m1 is T=m1g. The tension in the string is transmitted all the way round to m2. The total downward force on the pulley - and therefore also the tension in the spring - is F=2T=2m1g. (We don't need to know the tension in the string below m2.)

The PE stored in the spring is 12kx2=F22k=2(m1g)2k.

Answer 2

Answer:

Let the extension in spring be x

Then m1 moves down by 2x .  

By energy conservation m1gx=12kx2

Hence x= (4mg)/(k)

So PE is 8m2g2k

But the answer given as 2m2g2k

If m2 is attached to the ground then it doesn't really matter what value it has, provided that it is less than m1. The string or wire which attaches m2 to the ground will supply enough force to counteract the weight of m1. (If m2>m1 then there is no equilibrium.)

At equilibrium the tension in the string holding up m1 is T=m1g. The tension in the string is transmitted all the way round to m2. The total downward force on the pulley - and therefore also the tension in the spring - is F=2T=2m1g. (We don't need to know the tension in the string below m2.)

The PE stored in the spring is 12kx2=F22k=2(m1g)2