Question

$\rm{ \green{Hello \: Brainlians! }}$
Please help me to solve this Question.
$\bf{ \orange{Given \ Information}}$
Two blocks P and Q of equal masses 1 kg are placed on a rough inclined plane. Initially the block P is $\sf{\sqrt{2}}$ m behind the block Q. While moving down the incline, block P and Q experience a retarding force $\sf{\sqrt{2 \ N}}$ and $\sf{\dfrac{3}{\sqrt{2}}}$ N respectively.
1. If the two blocks are released simultaneously, then find the time taken by the blocks to come on the same line on the inclined plane as shown in the figure.

2.Find the distance travelled by the blocks along the incline till they come on the same line on the inclined plane.
Please, do not spam. Best answer will be marked as Brainliest.
$\sf{ \purple{Bonne \ Chance!}}$
​

Answer 1

$\huge\bf{\underbrace{\gray{SOLUTION:-}}}$

☯︎ First of all, we need to calculate the value of "coefficient of friction betⁿ plane and block P" and "coefficient of friction betⁿ plane and block Q" .

✯ Frictional force acting on block P is,

$\huge\red\checkmark$ $\bf{F_{P}\:=\:N_P\:\times{\mu}_P\:}$

$\bf{:\implies\:F_{P}\:=\:m_{P}\:g\cos{45^{\degree}}\times{\mu}_P\:}$

☞︎︎︎ It is given that,

• $\bf\red{F_{P}}$ = $\bf{\sqrt{2}\:N}$

• $\bf\red{m_{P}}$ = 1 kg

• $\bf\red{g}$ = 10 m/s²

$\rm{:\implies\:\sqrt{2}\:=\:1\times{10}\times{\dfrac{1}{\sqrt{2}}}\times{\mu}_{P}\:}$

$\rm{:\implies\:{\mu}_{P}\:=\:\dfrac{\sqrt{2}\times\:\sqrt{2}}{10}\:}$

$\rm{:\implies\:\mu_{P}\:=\:\dfrac{2}{10}\:}$

$\bf\green{:\implies\:\mu_{P}\:=\:0.2\:N\:}$

✯ Again frictional force acting on block Q is,

$\huge\red\checkmark$ $\bf{F_{Q}\:=\:N_Q\:\times{\mu}_Q\:}$

$\bf{:\implies\:F_{Q}\:=\:m_{Q}\:g\cos{45^{\degree}}\times{\mu}_Q\:}$

☞︎︎︎ It is given that,

• $\bf\red{F_{Q}}$ = $\bf{\dfrac{3}{\sqrt{2}}\:N}$

• $\bf\red{m_{Q}}$ = 1 kg

• $\bf\red{g}$ = 10 m/s²

$\rm{:\implies\:\dfrac{3}{\sqrt{2}}\:=\:1\times{10}\times{\dfrac{1}{\sqrt{2}}}\times{\mu}_{Q}\:}$

$\rm{:\implies\:\mu_{Q}\:=\:\dfrac{\sqrt{2}\times\:\dfrac{3}{\sqrt{2}}}{10}\:}$

$\rm{:\implies\:\mu_{Q}\:=\:\dfrac{3}{10}\:}$

$\bf\green{:\implies\:\mu_{Q}\:=\:0.3\:N\:}$

☯︎ Now, we have to calculate the acceleration of each block . For acceleration see the attachment figure-2 .

✞︎ Hence relative acceleration is,

$\huge\red\checkmark$ $\bf\purple{a_{relative}\:=\:a_P\:-\:a_Q\:}$

$\rm{:\implies\:a_{rel}\:=\:g\:(\sin{\theta}\:-\:\mu_{P}\:\cos{\theta})\:-\:g\:(\sin{\theta}\:-\:\mu_{Q}\:\cos{\theta})\:}$

$\rm{:\implies\:a_{rel}\:=\:g\:(\sin{\theta}\:-\:\mu_{P}\:\cos{\theta}\:-\:\sin{\theta}\:+\:\mu_{Q}\:\cos{\theta})\:}$

$\rm{:\implies\:a_{rel}\:=\:g\:(-\:\mu_{P}\:\cos{\theta}\:+\:\mu_{Q}\:\cos{\theta})\:}$

$\rm{:\implies\:a_{rel}\:=\:g\:\Big(-0.2\:\cos{45^{\degree}}\:+\:0.3\:\cos{45^{\degree}}\Big)\:}$

$\rm{:\implies\:a_{rel}\:=\:10\:\Big(-0.2\times{\dfrac{1}{\sqrt{2}}}\:+\:0.3\times{\dfrac{1}{\sqrt{2}}}\Big)\:}$

$\rm{:\implies\:a_{rel}\:=\:10\:\Big(-\dfrac{0.2}{\sqrt{2}}\:+\:\dfrac{0.3}{\sqrt{2}}\Big)\:}$

$\rm{:\implies\:a_{rel}\:=\:10\:\Big(\dfrac{0.3\:-\:0.2}{\sqrt{2}}\Big)\:}$

$\rm{:\implies\:a_{rel}\:=\:10\times{\dfrac{0.1}{\sqrt{2}}}\:}$

$\bf\green{:\implies\:a_{relative}\:=\:\dfrac{1}{\sqrt{2}}\:m/s^2\:}$

➪ Initially, block's are in rest .

$\green\bigstar\:\bf\orange{S_{rel}\:=\:\dfrac{1}{2}\:a_{rel}\:t^2\:}$

• $\bf\red{S_{rel}}$ = $\bf{\sqrt{2}\:m}$

$\rm{:\implies\:\sqrt{2}\:=\:\dfrac{1}{2}\times{\dfrac{1}{\sqrt{2}}}\times{t^2}\:}$

$\rm{:\implies\:\sqrt{2}\:=\:\dfrac{1}{2\sqrt{2}}\times{t^2}\:}$

$\rm{:\implies\:t^2\:=\:\sqrt{2}\times{2\sqrt{2}}\:}$

$\rm{:\implies\:t^2\:=\:4\:}$

$\rm{:\implies\:t\:=\:\sqrt{4}\:}$

$\bf\blue{:\implies\:t\:=\:2\:second}$

$\huge\red\therefore$ [1] If the two blocks are released simultaneously, then the time taken by the blocks to come on the same line on the inclined plane is "2 second" .

__________________________

✯ Distance moved by block P in this time period is,

$\purple\bigstar\:\bf\pink{S_{P}\:=\:\dfrac{1}{2}\:a_{P}\:t^2\:}$

$\rm{:\implies\:S_{P}\:=\:\dfrac{1}{2}\:\Big\{g\:(\sin{\theta}\:-\:\mu_{P}\:\cos{\theta})\Big\}\:t^2\:}$

$\rm{:\implies\:S_{P}\:=\:\dfrac{1}{2}\times{10}\:(\sin{45^{\degree}}\:-\:0.2\:\cos{45^{\degree}})\:\times{2^2}\:}$

$\rm{:\implies\:S_{P}\:=\:20\times{\dfrac{1}{\sqrt{2}}}\:(1\:-\:0.2)\:}$

$\rm{:\implies\:S_{P}\:=\:10\sqrt{2}\times\:0.8\:}$

$\bf\blue{:\implies\:S_{P}\:=\:8\sqrt{2}\:m}$

$\huge\red\therefore$ Distance moved by block P in this time period is "8√2 m" .

✯ Distance moved by block Q in this time period is,

$\purple\bigstar\:\bf\pink{S_{Q}\:=\:\dfrac{1}{2}\:a_{Q}\:t^2\:}$

$\rm{:\implies\:S_{Q}\:=\:\dfrac{1}{2}\:\Big\{g\:(\sin{\theta}\:-\:\mu_{Q}\:\cos{\theta})\Big\}\:t^2\:}$

$\rm{:\implies\:S_{Q}\:=\:\dfrac{1}{2}\times{10}\:(\sin{45^{\degree}}\:-\:0.3\:\cos{45^{\degree}})\:\times{2^2}\:}$

$\rm{:\implies\:S_{P}\:=\:20\times{\dfrac{1}{\sqrt{2}}}\:(1\:-\:0.3)\:}$

$\rm{:\implies\:S_{P}\:=\:10\sqrt{2}\times\:0.7\:}$

$\bf\blue{:\implies\:S_{P}\:=\:7\sqrt{2}\:m}$

$\huge\red\therefore$ Distance moved by block Q in this time period is "7√2 m" .

Answer 2

Given :

• Mass of blocks P and Q , M = 1 kg
• Intitally the block P is √2 m behind the block Q .
• While moving down the incline, block P and Q experience a retarding force $\sf{\sqrt{2}}$N and $\sf{\dfrac{3}{\sqrt{2}}}$ N

To Find :

1. If the two blocks are released simultaneously, then find the time taken by the blocks to come on the same line on the inclined plane as shown in the figure.
2. Find the distance travelled by the blocks along the incline till they come on the same line on the inclined plane.

Solution :

Before solving the question , First draw the free body diagram of given scenario.

1) We have to find the time taken by the blocks to come on the same line on the inclined plane. If the two blocks are released simultaneously.

Let the accelaration of block P and Q be $\sf\:a_1\:and\:a_2$

From the free body diagram :

Forces acting on Block P

$\sf\:Mg\sin45-F_1=Ma_1$

$\sf\implies\:a_1=\dfrac{Mg\sin\:45-F_1}{M}$

$\sf\implies\:a_1=g\sin\:45-F$

$\sf\implies\:a_1=10\times\dfrac{1}{\sqrt{2}}-\sqrt{2}$

$\sf\implies\:a_1=\dfrac{10-2}{\sqrt{2}}$

$\sf\implies\:a_1=\dfrac{8}{\sqrt{2}}$

$\sf\implies\:a_1=4\sqrt{2}$

Forces acting on Block Q

$\sf\:Mg\sin45-F_2=Ma_2$

$\sf\implies\:a_2=\dfrac{Mg\sin\:45-F_2}{M}$

$\sf\implies\:a_2=g\sin\:45-F_2$

$\sf\implies\:a_2=10\times\dfrac{1}{\sqrt{2}}-\dfrac{3}{\sqrt{2}}$

$\sf\implies\:a_2=\dfrac{10-3}{\sqrt{2}}$

$\sf\implies\:a_2=\dfrac{7}{\sqrt{2}}$

$\sf\implies\:a_2=3.5\sqrt{2}$

Now Acceleration of Q observed by P

$\sf\:a_{21}=a_2-a_1=0.5\sqrt{2}$

By equation of motion :

$\sf\triangle\:S=ut+\dfrac{1}{2}\times\:a_{21}t^2$

$\sf\sqrt{2}=0+\dfrac{1}{2}\times0.5\sqrt{2}t^2$

$\sf\:t^2=\dfrac{\sqrt{2}\times2}{0.5\times\sqrt{2}}$

$\sf\:t^2=4$

$\sf\:t=2\:sec$

Therefore , the time taken by the blocks to come on the same line on the inclined plane is 2 sec.

2) We have to find the distance travelled by the blocks along the incline till they come on the same line on the inclined plane.

Distance travelled by Block

$\sf\:S_1=ut+\dfrac{1}{2}\times\:a_{1}t^2$

Put the value of $\sf\:a_1=4\sqrt{2}$

$\sf\implies\:S_1=0+\dfrac{1}{2}\times\:4\sqrt{2}\times4$

$\sf\implies\:S_1=8\sqrt{2}m$

Similarly ,

Distance travelled by Block Q

$\sf\:S_2=ut+\dfrac{1}{2}\times\:a_{2}t^2$

Put the value of $\sf\:a_2=3.5\sqrt{2}$

$\sf\implies\:S_2=0+\dfrac{1}{2}\times\:3.5\sqrt{2}\times4$

$\sf\implies\:S_2=7\sqrt{2}m$

Hence ,the distance travelled by the block P and Q along the incline till they come on the same line on the inclined plane is 8√2 and 7√2 m.