A uniform plank of length L , cross sectional area A and Young's modulus Y , is resting over a smooth horizontal surface it is subjected to a horizontal force at its one end . Which of the following statements are not correct?

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$S t r e s s d e v e l o p e d i n p l a n k m a t e r i a l i s m a x i m u m a t t h e e n d a t w h i c h f o r c e i s a p p l i e d a n d d e c r e a s e l i n e a r l y t o z e r o a t t h e o t h e r e n d .$

$A u n i f o r m t e n s i l e s t r e s s i s d e v e l o p e d i n t h e p l a n k m a t e r i a l .$

$\sin c e p l a n k i s p u l l e d a t o n e e n d o n l y p l a n k s t a r t s t o a c c e l e r a t e a l o n g d i r e c t i o n o f t h e f o r c e . H e n c e n o s t r e s s i s d e v e l o p e d i n t h e p l a n k m a t e r i a l .$

Elastic potential energy stored in the plank is $\frac{F^{2} L}{6 A Y}$

answer is A, D.

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Detailed Solution

Let mass of the plank is $M$ and force $F$ is applied as shown. Then at a distance $x$ from the other end

$T = \frac{M}{L} \times x \times \frac{F}{M} = \frac{x}{L} F$

Hence stress is $σ_{x} = \frac{T}{A} = \frac{F}{A L} \times x$ .

$∴$ Option (1) is correct.

$u_{x} = \frac{{σ_{x}}^{2}}{2 Y} \Rightarrow d U_{x} = u_{x} (A . d x) \Rightarrow U = \int_{0}^{L} A u_{x} d x = \frac{F^{2} L}{6 A Y}$ . So option $(4) i s r i g h t .$