Question 255: Work, Energy and Power & Circular Motion

Question

A block is fastened to a horizontal spring. The block is pulled to a distance $$x=10 \mathrm{~cm}$$ from its equilibrium position (at $$x=0$$) on a frictionless surface from rest. The energy of the block at $$x=5$$ $$\mathrm{cm}$$ is $$0.25 \mathrm{~J}$$. The spring constant of the spring is ___________ $$\mathrm{Nm}^{-1}$$

Accepted Answer

Spring energy at x = 10 cm,
$$\mathrm{U}_{\mathrm{i}}  =\frac{1}{2} \mathrm{kx}_0^2 $$
Energy of the block at x = 10,
$$\mathrm{~K}_{\mathrm{i}}  =0$$
Spring energy at x = 5 cm,
$$\mathrm{U}_{\mathrm{f}}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{x}_0}{2}\right)^2 $$
Energy of the block at x = 5, (which is only kinetic energy, no potential energy of block presents as block is not moving in the vertical direction)
$$
 \mathrm{~K}_{\mathrm{f}}=0.25 \mathrm{~J} $$
Applying energy conservation law,
Initial energy of Spring + Initial energy of Block = Final energy of Spring + Final energy of Block
$$
 \frac{1}{2} \mathrm{kx}_0^2+0=\frac{1}{2} \mathrm{k} \frac{\mathrm{x}_0^2}{4}+0.25 $$
$$
 \frac{1}{2} \mathrm{kx}_0^2 \frac{3}{4}=\frac{1}{4} $$
$$
 \frac{1}{2} \mathrm{k} \frac{3}{100}=1 \Rightarrow \mathrm{k}=\frac{200}{3} \mathrm{~N} / \mathrm{m} $$
$$
 =67 \mathrm{~N} / \mathrm{m}

$$

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type