Consider the system shown below, with two equal masses m and a spring with spring constant K. The coefficient of friction between the left mass and horizontal table is $μ = \frac{1}{4}$ , and the pulley is frictionless. The string connecting both the blocks is massless and inelastic. The system is held with the spring at its unstretched length and then released. The extension in spring when the masses come to momentary rest for the first time is $x_{1}$ and in the absence of friction the spring is elongated by $x_{2}$ before reaching mean position. Calculate $\frac{x_{1}}{x_{2}}$ .

Moderate

Solution

from work energy theorem, the masses stop when total work done on them is zero.
$\begin{array}{l} w = m g x - \frac{1}{2} k x^{2} - μ m g x = 0 \\ \times = \frac{2 m g}{k} (1 - μ) = \frac{3 m g}{2 k} \end{array}$
In the absence of friction if the spring is elongated by $x_{2}$ before reaching the mean position then $x_{2} = \frac{m g}{k}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME