Work Power Energy Question 9
Question 9 - 01 February - Shift 2
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 $Nm^{-1}$.
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Answer: (50)
Solution:
Formula: Potential Energy
$U_i=\frac{1}{2} kx_0^{2}$
$K_i=0$
$U_f=\frac{1}{2} k(\frac{x_0}{2})^{2}$
$K_f=0.25 J$
$\frac{1}{2} kx_0^{2}+0=\frac{1}{2} k \frac{x_0^{2}}{4}+0.25$
$\frac{1}{2} kx_0^{2} \frac{3}{4}=\frac{1}{4}$
$\frac{1}{2} k \frac{3}{100}=1 \Rightarrow k=\frac{200}{3} N / m$
$=67 N / m$
