Question

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 distance

when released from rest?

[Take gravitational acceleration as 10 m/s2]​

Answer 1

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

Answer 2

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:}}}

†Tofind:

Time taken to travel

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

†Taken:

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

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

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} }s=ut+

2

1

at

2

Where,

s = Distance travelled

ut = Initial velocity

a = Acceleration

t = Time taken

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

†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} }a⟹

m1+m2

(m1−g)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:}}}

†Solution:

First , acceleration:

\bold\purple{Taken,a=\frac{(m1-g)m2}{m1+m2}}Taken,a=

m1+m2

(m1−g)m2

\bold{a=\frac{(20kg-10m/s)10}{20kg+10kg}}a=

20kg+10kg

(20kg−10m/s)10

\bold{a=\frac{10×10}{30}}a=

30

10×10

\bold{a=\frac{10}{3}}a=

3

10

___________________________________

Now , find the time taken:

\bold\purple{Taken,s=ut+\frac{1}{2}at²}Taken,s=ut+

2

1

at²

\bold{0.6m=0+\frac{1}{2}×\frac{10}{3}×t²}0.6m=0+

2

1

×

3

10

×t²

\bold{t²=\frac{3.6}{10}}t²=

10

3.6

\bold{t²=\frac{36}{100}}t²=

100

36

\bold{t = \sqrt{ \frac{36}{100} } }t=

100

36

\bold{t = \frac{6}{10} }t=

10

6

\bold{t=0.6}t=0.6

\huge{\underline{\dag{Answer:}}}

†Answer:

So , the time taken by the first block is 0.6 seconds .

\huge{\underline{\dag{Extra\:information:}}}

†Extrainformation:

First equation of motion:

\bold{Vf=Vi+at}Vf=Vi+at

Where,

Vf = Final velocity

Vi = Initial velocity

a = Acceleration

t = Time

Third equation of motion:

\bold{v²-u²=at}v²−u²=at

Where,

v² = Final velocity

u² = Initial velocity

a = Acceleration

t = Time

Fourth equation of motion:

\bold{v²=u²+2as}v²=u²+2as

Where,

v² = Final velocity

u² = Initial velocity

a = Acceleration

HOPE IT HELPS YOU