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BITSAT 2015 Paper

Section: Physics

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Question : 10 of 150

Marks: +1, -0

An elastic string of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is

1/2 $K y^2$
1/2 Ky(2x + y)
1/2 K( $x^2 + y^2$ )
1/2 k $(x + y)^2$

Solution:

In the string elastic force is conservative in nature.

∴ W = - ΔU

Work done by elastic force of string,

W = -

(U_F-U_i)

=

U_i-U_F

W =

\frac{1}{2} k x^2 - \frac{k}{2} (x+y)^2

=

\frac{1}{2} k x^2 - \frac{1}{2} k (x^2+y^2+2xy)

=

\frac{1}{2} k x^2 - \frac{1}{2} k y^2

-

\frac{1}{2} k x^2 - \frac{1}{2} k (2xy)

= - kxy -

\frac{1}{2} k y^2

Therefore, the work done against elastic force

W_{\text{external}}

= - W = 1/2 Ky(2x + y)

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