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 $025 J$. The spring constant of the spring is _____$Nm ^{-1}$.

Answer 1

Correct Answer: 50

The correct answer is 50.

$U_f=\frac{1}{2}k{\left(\frac{x_0}{2}\right)}^2$
$K_f=0.25J$
$\frac{1}{2}k{x}_0^2+0=\frac{1}{2}k\frac{x_0^2}{4}+0.25$
$\frac{1}{2}k{x}_0^2\frac{3}{4}=\frac{1}{4}$
$\frac{1}{2}k\frac{3}{100}=1\Rightarrow k=\frac{200}{3}N/m$
$=67N/m$

Answer 2

At first the block is pulled to a distance \(X_0 = 10 \ cm\) (extreme position) from its \(X=0\) (equilibrium position).
Initial Potential Energy of the block is \(U_i=\frac 12kX_0^2\) and 
Initial Kinetic Energy of the block is \(k_i=0\)
Given, the block is at the distance of \(5\ cm\).

Final Potential Energy of the block is \(U_f=\frac 12k(\frac {X_0}{2})^2\) and
Given, Kinetic Energy of the block is \(K_f= 0.25\ J\)
To solve this problem, we can use the law of conservation of Energy. The total energy of the system remains constant throughout the motion.
Therefore, as the Total Energy of the block will be conserved, so
\(U_i+K_i=U_f+K_f\)

\(\frac 12kX_0^2+0=\frac 12k(\frac {X_0}{2})2+0.25\)

\(\frac12kX_0^2(\frac 34) = \frac 14\)

\(\frac 12k(\frac {3}{100})=1\)

\(k=\frac {220}{3} Nm^{-1}\)
\(k =67\ Nm^{-1}\)

Therefore, the Spring constant of the spring is 67 N/m.