Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, the maximum elongation of the spring is

Question

Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, the maximum elongation of the spring is (A) (m g/k) (B) (2 m g/k) (C) (3 m g/k) (D) (4 m g/k)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/360a9ab1c81713e6492373d509aa5436-.png" /> Δ K+Δ Ug+Δ Us=0 0+(-m g x max )+(1/2) k x max 2=0 ⇒ (1/2) k x max 2=m g x max ⇒ x max =((2 m g/k)).

Q. Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, the maximum elongation of the spring is

$\frac{m g}{k}$

$\frac{2 m g}{k}$

$\frac{3 m g}{k}$

$\frac{4 m g}{k}$

$Δ K + Δ U_{g} + Δ U_{s} = 0$
$0 + (- m g x_{m a x}) + \frac{1}{2} k x_{m a x}^{2} = 0$
$\Rightarrow \frac{1}{2} k x_{m a x}^{2} = m g x_{m a x}$
$\Rightarrow x_{m a x} = (\frac{2 m g}{k})$ .

Q. Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, the maximum elongation of the spring is

kmg​

k2mg​

k3mg​

k4mg​

ΔK+ΔUg​+ΔUs​=0 0+(−mgxmax​)+21​kxmax2​=0 ⇒21​kxmax2​=mgxmax​ ⇒xmax​=(k2mg​).

$\frac{m g}{k}$

$\frac{2 m g}{k}$

$\frac{3 m g}{k}$

$\frac{4 m g}{k}$

$Δ K + Δ U_{g} + Δ U_{s} = 0$
$0 + (- m g x_{m a x}) + \frac{1}{2} k x_{m a x}^{2} = 0$
$\Rightarrow \frac{1}{2} k x_{m a x}^{2} = m g x_{m a x}$
$\Rightarrow x_{m a x} = (\frac{2 m g}{k})$ .