The system shown in the figure is released fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $x$ be the extension in the spring when the $2\, kg$ block leaves contact with the ground. At this instant,the spring force must balance the w

Vedclass · Accepted Answer

$2\sqrt{2}\, m/s$

Vedclass · Answer

(B) Let $x$ be the extension in the spring when the $2\, kg$ block leaves contact with the ground. At this instant,the spring force must balance the weight of the $2\, kg$ block.$Kx = m_2g$$40 	imes x = 2 	imes 10$$x = \frac{20}{40} = 0.5\, m$Now,we apply the principle of conservation of mechanical energy for the $5\, kg$ block as it moves down by a distance $x = 0.5\, m$. The loss in gravitational potential energy of the $5\, kg$ block is equal to the gain in elastic potential energy of the spring plus the gain in kinetic energy of the $5\, kg$ block.$m_1gx = \frac{1}{2}Kx^2 + \frac{1}{2}m_1v^2$$5 	imes 10 	imes 0.5 = \frac{1}{2} 	imes 40 	imes (0.5)^2 + \frac{1}{2} 	imes 5 	imes v^2$$25 = 20 	imes 0.25 + 2.5v^2$$25 = 5 + 2.5v^2$$20 = 2.5v^2$$v^2 = \frac{20}{2.5} = 8$$v = \sqrt{8} = 2\sqrt{2}\, m/s$

The system shown in the figure is released from rest. The pulley and spring are massless,and friction is absent everywhere. The speed of the $5\, kg$ block when the $2\, kg$ block leaves contact with the ground is: (Take the force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2$)

Similar Questions

The energy $(E)$,angular momentum $(L)$,and universal gravitational constant $(G)$ are chosen as fundamental quantities. The dimensions of the universal gravitational constant in the dimensional formula of Planck's constant $(h)$ is

Evaluate $\int_{-2}^1 f(x) dx$,where $f(x)= \begin{cases} 1-2x, & x \leq 0 \\ 1+2x, & x > 0 \end{cases}$

The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then,the equation of this circle,which also passes through $(5,7)$,is

$A$ polaroid is placed at $45^{\circ}$ to an incoming light of intensity $I_0$. Now,the intensity of light passing through the polaroid after polarization would be:

In the reaction,the major product formed is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module