A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at x = 5

Question

A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at x = 5 cm is 0 . 25 J . The spring constant of the spring is ______ N m - 1 .

TestHub · Accepted Answer

At first the block is pulled to a distance $x_{0} = 10 cm$ (extreme position).

Potential energy of block is $U_{i} = \frac{1}{2} k x_{0}^{2}$ and kinetic energy os block, $K_{i} = 0$ .

Now, when the block is at distance $5 cm$ .

Potential energy is $U_{f} = \frac{1}{2} k {(\frac{x_{0}}{2})}^{2}$ and kinetic energy is $K_{f} = 0.25 J$

Total energy of block will be conserved, so

$\frac{1}{2} k x_{0}^{2} + 0 = \frac{1}{2} k \frac{x_{0}^{2}}{4} + 0.25$

$\Rightarrow \frac{1}{2} k x_{0}^{2} \frac{3}{4} = \frac{1}{4}$

$\Rightarrow \frac{1}{2} k \frac{3}{100} = 1 \Rightarrow k = \frac{200}{3} N m^{- 1}$

$= 67 N m^{- 1}$

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

Doubts & Discussion