Question

14. Initially spring is relaxed when m1, is released from rest. Calculate for what minimum values
block of mass m will just leave the contact with surface ? Give answer in terms of
m/m1​

Answer 1

Answer:

b unit er result kobe dibe na pakka bagh the

Answer 2

Concept:

In any system the energy is conserved, that is the energy can neither be created nor destroyed.

Given:

A pulley, Spring, and the two masses m and m₁ are attached.

Find:

The minimum value of m in terms of m₁.

Solution:

From the spring and mass m, balance,

mg = Kx, where K is the spring constant and x is the extension in the string.

And, the kinetic energy, KE = 1/2 Kx²

From the energy balance on the mass m, here, x is the maximum displacment.

Potential energy = Kinetic energy

m₁gx = 1/2 Kx²

From above mg = Kx,

m₁gx = 1/2 Kx(x)

m₁gx = 1/2 (mg)(x)

Canceling the common terms,

m₁ = 1/2m

m/m₁ = 2

Hence, the minimum values block of mass m for which it will just leave contact with the surface, the value of m/m₁ is 2.

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