Question #306301

A catapult consists of a rubber of unstretched length, 0.20m. The rubber is stretched by 0.10m by a tension of 50N. A stone of mass 0.06Kg is projected vertically upwards from the catapult after the rubber has been stretched to a total length of 0.35m. Calculate the energy stored in the stretched catapult and determine the maximum height reached by the stone.

1
Expert's answer
2022-03-08T08:55:18-0500

Explanations & Calculations


  • Apply Hook's law to the rubber to find the elastic constant,

F=kx50N=k(0.10m)k=500Nm1\qquad\qquad \begin{aligned} \small F&=\small kx\\ \small 50\,N&=\small k(0.10\,m)\\ \small k&=\small 500\,Nm^{-1} \end{aligned}

  • Once the rubber is stretched to 0.35m its extension is (0.35-0.20 =)0.15m.
  • Now, the stored elastic energy (E.E) converts into potential energy (P.E) as a whole.
  • We can consider the stretched position as the level of zero potential energy and apply the conservation of mechanical energy for this happening.

E.E+P.E=P.E12kx2+0=mg(H+x)0.5×500Nm1×0.152m2=0.06kg×9.8ms2×(H+0.15)H=9.42m\qquad\qquad \begin{aligned} \small E.E+P.E&=\small P.E\\ \small \frac{1}{2}kx^2+0&=\small mg(H+x)\\ \small 0.5\times500\,Nm^{-1}\times0.15^2m^2&=\small 0.06kg \times9.8ms^{-2}\times(H+0.15)\\ \small H&=\small 9.42\,m \end{aligned}


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