IIT JEE , AIEEE Forum - Physics , Mechanics

bantu

In the figure given,find the work done in slowly pulling down ' M2 ' by 8 cm.

Given: k1 =1500 N/m , k2 =500 N/m, M1=2kg, M2=1kg . (g=10 m/s^2).

(answer = 1.2 J)

the figure is: a spring k1 suspended from wall with a mass M1 attached at the other end ,which is then continuedly attached with another spring k2attached with mass M2.

amey

Solve it using conservtion of energy principle

bantu

i had tried but not getting answer and have doubtof solution

konichiwa2x

I think question meant : Find the work done in slowly pulling down the system by 8cm. Only then, I am getting the solution you had provided.

Let a₁ and a₂ be the extensions in the springs at equilibirum.

Start by drawing the FBDS:

m₂g = k₂a₂ --- (1)

k₁a₁= m₁g + k₂a₂ --- (2)

a₂ = 0.02m and a₁ = 0.02m.

Let x₁ and x₂ be additional elongations causes by pulling m₂ by L = 8cm. The additional forces on m1 are equal and are in opposite directions.

Thus you can infer that k₁x₁ = k₂x₂ --- (3)

Also, use the constraint x₁ + x₂ = L --- (4)

Solve (3) and (4) to obtain the values of x₁ and x₂ as 0.06m and 0.02m respectively.

Apply work energy theorem now. Initial and final kinetic energy is 0.

W_g + W_p + W_s = 0 --- (5) where W_pis the work done by pulling force.

W_s = U₁ - U₂ where U₂ = 1/2 k₁(a₁ + x₁)² + 1/2 k₂(a₂ + x₂)²

Put in the values, and you get W_p= 1.2 J.

By the way, if the above is not clear, you might want to refer "Concepts of Physics - H.C Verma). I remember seeing a quite similar problem in it long back.

cheers.

sudeep.kumar

Nice job konichiwa2x (is it ur name??)... a salute for this one...

And yes.. pulling slowly m₂ down will result into the system being pulled down.. isnt it??

konichiwa2x

thank you sir. no, my real name is Abhijith.

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