Two blocks of mass 20 kg and 10 kg are connected by a massless string

1.	Two blocks of mass 20 kg and 10 kg are connected by a massless string as shown in the figure. How much time will 20 kg block take to travel 0,6 m distancewhen released from rest?[Take gravitational acceleration as 10 m/s2]
Answer» Answer: $\huge{\underline{\dag{Given:}}}$ Mass of two blocks which are connected to massless string: >>> 20 kg and 10 kg DISTANCE travel by the 20 kg block: >>> 0.6 m Acceleration due to gravity: >>> 10 m/s² INITIAL velocity: >>> 0 $\huge{\underline{\dag{To\:find:}}}$ Time TAKEN to travel $\huge{\underline{\dag{Taken:}}}$ First find the acceleration of the both block with the massless string: $a\implies \bold { \frac{m1 - m2}{m1 + m2} }$ Where, a = Acceleration m1 = Mass of the FIRST block m2 = Mass of the second block And then use the second equation of MOTION to find the time taken: $\bold{ s =ut + \frac{1}{2} a {t}^{2} }$ Where, s = Distance travelled ut = Initial velocity a = Acceleration t = Time taken $\huge{\underline{\dag{Concept:}}}$ To find the acceleration subtract mass of 1 st block with the acceleration due to gravity like this : $a \implies \bold{ \frac{(m1 - g)m2}{m1 + m2} }$ Where, a = Acceleration m1 = Mass of the first block m2 = Mass of the second block g = Acceleration due to gravity $\huge{\underline{\dag{Solution:}}}$ First , acceleration: $\bold\purple{Taken,a=\frac{(m1-g)m2}{m1+m2}}$ $\bold{a=\frac{(20kg-10m/s)10}{20kg+10kg}}$ $\bold{a=\frac{10×10}{30}}$ $\bold{a=\frac{10}{3}}$ ___________________________________ Now , find the time taken: $\bold\purple{Taken,s=ut+\frac{1}{2}at²}$ $\bold{0.6m=0+\frac{1}{2}×\frac{10}{3}×t²}$ $\bold{t²=\frac{3.6}{10}}$ $\bold{t²=\frac{36}{100}}$ $\bold{t = \sqrt{ \frac{36}{100} } }$ $\bold{t = \frac{6}{10} }$ $\bold{t=0.6}$ $\huge{\underline{\dag{Answer:}}}$ So , the time taken by the first block is 0.6 seconds . $\huge{\underline{\dag{Extra\:information:}}}$ First equation of motion: $\bold{Vf=Vi+at}$ Where, Vf = Final velocity Vi = Initial velocity a = Acceleration t = Time Third equation of motion: $\bold{v²-u²=at}$ Where, v² = Final velocity u² = Initial velocity a = Acceleration t = Time Fourth equation of motion: $\bold{v²=u²+2as}$ Where, v² = Final velocity u² = Initial velocity a = Acceleration HOPE IT HELPS YOU

Two blocks of mass 20 kg and 10 kg are connected by a massless string as shown in the figure. How much time will 20 kg block take to travel 0,6 m distancewhen released from rest?[Take gravitational acceleration as 10 m/s2]

Answer»

Answer:

$\huge{\underline{\dag{Given:}}}$

Mass of two blocks which are connected to massless string:

>>> 20 kg and 10 kg

DISTANCE travel by the 20 kg block:

>>> 0.6 m

Acceleration due to gravity:

>>> 10 m/s²

INITIAL velocity:

>>> 0

$\huge{\underline{\dag{To\:find:}}}$

Time TAKEN to travel

$\huge{\underline{\dag{Taken:}}}$

First find the acceleration of the both block with the massless string:

$a\implies \bold { \frac{m1 - m2}{m1 + m2} }$

Where,

a = Acceleration

m1 = Mass of the FIRST block

m2 = Mass of the second block

And then use the second equation of MOTION to find the time taken:

$\bold{ s =ut + \frac{1}{2} a {t}^{2} }$

Where,

s = Distance travelled

ut = Initial velocity

a = Acceleration

t = Time taken

$\huge{\underline{\dag{Concept:}}}$

To find the acceleration subtract mass of 1 st block with the acceleration due to gravity like this :