Question

Frictional force acting on the mass $2m$ when it finally comes to rest is

• A. $16N$
• B. $8N$
• C. $12N$
• D. $zero$

Answer 1

Answer: (C)

$12N$

Due to weight of block m, let spring gets elongated by distance $x$.

As one end of the rope is fixed , the block of mass m will go down by distance $2x$.

Given:

Initial Velocity $u=0$ and at maximum extension

spring will come to rest for a moment so final Velocity $v=0$.

change in kinetic energy $\Delta KE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

$\Delta KE=0$

From work energy theorem , it is known

$\Delta KE=W_g+W_f+W_s$, where $W_g$ is work done by gravity,

$W_f$ is work done by friction on block of mass $2m$ and $W_s$ is work done by spring.

$W_g=mg\times 2x=1\times 10\times 2x=20x$

$W_s=-\frac{1}{2}k{x}^2=-0.5\times 100x^2=-50x^2$

$W_f=-\mu 2mgx=-0.8\times 2\times 10x=-16x$

putting values in

$\Delta KE=W_g+W_f+W_s$

$0=20x-50x^2-16x$

$50x^2-4x=0$

$x=0.08m$

$x=80cm$

Let $T$ be the tension in spring attached with mass $m$.

so tension in string attached with mass $2m$ be $T+T=2T$.

Mass $m$ is in rest also, net force acting on it =0

assuming upward direction to be +

$T-mg=0$

$T=mg$

$T=1\times 10=10N$

Force acting on block of mass $2m$,

$2T\text{in}\leftarrow$ ,

$Kx\text{in}\to$ and friction force $F_{fr}$

$F_{fr}\text{in}\to$

Mass $2m$ is in rest, net force acting=0

assuming right direction to be +

$Kx+F_{fr}-2T=0$

putting values

$100\times 0.08+F_{fr}-2\times 10=0$

$8+F_{fr}-20=0$

$F_{fr}=12N$