Two blocks of equal masses are connected with | vedclass

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

Question

(B) Let the mass of each block be $m$. The system moves with a common acceleration $a = \frac{F}{2m}$.In the frame of the centre of mass, the blocks a

Vedclass · Accepted Answer

$10.4$

Vedclass · Answer

(B) Let the mass of each block be $m$. The system moves with a common acceleration $a = \frac{F}{2m}$.In the frame of the centre of mass, the blocks are at rest at the moment of maximum extension. The pseudo force acting on each mass is $F_p = ma = \frac{F}{2}$.Let $x_1$ and $x_2$ be the displacements of the two blocks from their initial positions in the centre of mass frame. The total extension of the spring is $x = x_1 + x_2$.The work done by the external force and pseudo forces in the centre of mass frame is equal to the change in potential energy of the spring:$W = \frac{1}{2} k x^2$In the centre of mass frame, the work done by the effective forces is:$(F - ma)x_1 + (ma)x_2 = \frac{1}{2} k (x_1 + x_2)^2$Substituting $a = \frac{F}{2m}$:$(F - \frac{F}{2})x_1 + (\frac{F}{2})x_2 = \frac{1}{2} k (x_1 + x_2)^2$$\frac{F}{2}(x_1 + x_2) = \frac{1}{2} k (x_1 + x_2)^2$$x_1 + x_2 = \frac{F}{k}$Given $F = 10 \,N$ and $k = 2500 \,N/m$, the maximum extension is:$x_{max} = \frac{10}{2500} \,m = 0.004 \,m = 0.4 \,cm$The maximum distance between the blocks is the sum of the natural length and the maximum extension:$d_{max} = 10 \,cm + 0.4 \,cm = 10.4 \,cm$.

Similar Questions

In the arrangement shown in the figure,if the blocks of masses $m$ and $2m$ are released from rest,find the tension in the string. (Given: $\mu_1 = 2/3$ for mass $m$,$\mu_2 = 1/3$ for mass $2m$,the string is massless and inextensible,and the pulley is frictionless.)

$A$ person with a machine gun can fire $50 \ g$ bullets with a velocity of $240 \ m/s$. $A$ $60 \ kg$ tiger moves towards him with a velocity of $12 \ m/s$. In order to stop the tiger in its track,the number of bullets the person must fire towards the tiger is:

Two observers are moving at a constant velocity $v$ relative to each other along a straight line. They observe a block of mass $m$ moving over a distance $l$ on a rough surface. Which of the following quantities will be the same for both observers?

What is the acceleration of the block and trolley system shown in the figure,if the coefficient of kinetic friction between the trolley and the surface is $0.04$? What is the tension in the string? (Take $g = 10\; m s^{-2}$). Neglect the mass of the string.

Vedclass Test Series

Exam Paper Generator

Online Exam Module