Question

A block of mass ${}^{\prime }{M}^{\prime }$ area of cross-section ${}^{\prime }{A}^{\prime }$ and length ${}^{\prime }{l}^{\prime }$ is placed on smooth horizontal floor. A force ${}^{\prime }{F}^{\prime }$ is applied on the block as shown. If ${}^{\prime }{y}^{\prime }$ is young modulus of material, then total extension in the block will be:

• A. $\frac{F\ell }{AY}$
• B. $\frac{F\ell }{2AY}$
• C. $\frac{F\ell }{3AY}$
• D. $cannotextend$

Answer 1

The correct option is B $\frac{F\ell }{2AY}$

Acceleration $a=\frac{F}{m}$

then,

$T=\frac{mx}{l}\times \frac{F}{m}=\frac{Fx}{l}$

Extension in $dx$ element

$-d\delta =\frac{Tdx}{Ay}=\frac{Fxdx}{lAy}$

Total extension ,

$\delta ={\int }_0^l\frac{Fxdx}{lAy}$

$\delta =\frac{Fl}{2Ay}$