Physics - W.E.P. Question with Solution | TestHub

If$l$is the stretched length of the spring

$\frac{\text{d}}{l}={\text{cos 37}}^{\text{o}}=\frac{4}{5}\text{,}\text{i.e.,}l=\frac{5}{4}\text{d}$

$\text{So, the stretch}\text{y}=l-\text{d}=\frac{5}{4}\text{d}-\text{d}=\frac{\text{d}}{4}$

$\text{and}\text{AB}=\text{h}=l{sin\; 37}^{\text{o}}=\frac{5}{4}\text{d}\times \frac{3}{5}=\frac{3}{4}\text{d}$

Now taking point$\mathrm{B}$as a reference level and applying the law of conservation of Energy between$\mathrm{A}$and$\mathrm{B}$,

${\mathrm{E}}_{\mathrm{A}} = {\mathrm{E}}_{\mathrm{B}}$

$\text{mgh}+\frac{1}{2}{\text{ky}}^2+0=0+0+\frac{1}{2}{\text{mv}}^2$

$\left[\text{as for B, h}=0\text{and y}=0\right]$

$\text{or}\frac{3}{4}\text{mgd}+\frac{1}{2}\text{k}{\left(\frac{\text{d}}{4}\right)}^2=\frac{1}{2}{\text{mv}}^2$

$\left[\text{as for A, h}=\frac{3}{4}\text{d}\text{and y}=\frac{1}{4}\text{d}\right]$

$\text{or}\text{v}=\text{d}\sqrt{\frac{\text{3g}}{\text{2d}}+\frac{\text{k}}{\text{16m}}}$