Springs - Conservation or Balancing Forces??

A block of mass m is lowered from natural length position of spring slowly by an external agent to equillibrium position .The extension present in the mass less spring was asked from 2 students

Student A : 1/2 kx2 = mgx

=> x = 2mg/k

Student B : mg = kx

=> x = mg/k

(A) Only A is correct

(B) Only B is correct

(C) Both correct

(D) Both wrong

13 Answers

19
Debotosh.. ·

discussed several times earlier !

4
UTTARA ·

I never noticed it before!!

Anyways What's the answer??

106
Asish Mahapatra ·

(B) .. think of the importance of the word "SLOWLY"

1
bhargavahariom ·

the answer is (B) becausenet force in d situation is 0 but non0 in(A)

6
Kalyan IIT-K Beware I'm coming ·

mg is an external force??m getting confused wid dat....

6
Kalyan IIT-K Beware I'm coming ·

oh i got it.... g is an external force....bt wat is wrong wid the 1st option??

1
Maths Musing ·

1 / 2 kx^2 ≠mgx

we have to consider the total displacement of com , { to conserve energy ,} which is x / 2 , not x .

so 1 / 2 k x^2 = m g x / 2

or x = mg / k

1
Bicchuram Aveek ·

The word "slowly" means that at each position an equilibrium is set up. thus in such a quasi-static process the 2nd option holds true.

1
Bicchuram Aveek ·

And Soumya.....it has nothing the hell to do with COM.

1
Arka Halder ·

Yes.Soumya, couldnt quite understand ur reasoning about the centre of mass.

23
qwerty ·

student A fogot the work done by external agent ??? or wat ??

6
Kalyan IIT-K Beware I'm coming ·

cud anyone pls tell me wat the hell is wrong in the 1st option.....???

1
Bicchuram Aveek ·

Yes kalyan...u conserve energy whn the entire system is in equilibrium not STATIC. if the word "slowly" hadn't been der...the ist option would have held true.

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