Question

In the shown diagram m${}_1$ = m${}_2$ = 4 kg and m${}_3$ = 2 kg.Coefficient of friction between m${}_1$ and m${}_2$ is 0.5. The mass m${}_1$ is given a velocity v and it just stops at the other end of the mass m${}_2$ in 2 sec. Let a${}_1$, a${}_2$ and a${}_3$ be the acceleration m${}_1$, m${}_2$ and m${}_3$ respectively, then:

• A. For t<1sec,,
• B. For t<1sec,,
• C. the value of v is 5 m/s
• D. For t<1sec,

Answer 1

Answer: (B), (C), (D)

The correct options are
B

For t<1sec,,

C

the value of v is 5 m/s

D

For t<1sec,

Okay so let's imagine how the motion will look like.

m${}_3$ will move down. m${}_2$ will slide rightwards m${}_3$ and m${}_2$ are connected by inextensible massless string so acceleration of these 2 are going to be same for sure

$\Rightarrow$ a${}_2$ = a${}_3$ = a

Given m${}_1$ is sliding on m${}_2$ in leftward direction with respect to m${}_2$

So friction on m${}_1$ will be backward or is rightward direction and according to 3rd law, equal force will be on m${}_2$ in leftward direction.

Let's draw free body diagram:

$m_{3}g-T=m_{3}a$

20 - T=2a .............(i)

T - $f_r$=$m_{2}a$

$f_r=\mu N=\mu m_{1}g=\frac{1}{2}\times 40=20N$

T - 20=4a ...........(ii)

From (I) $\&$ (II) a${}_2$ = a${}_3$ = a = 0

So m${}_2$ $\&$ m${}_3$ won't move till the block m${}_1$ is in relative motion with respect to m${}_2$

N=$m_{1}g$

$fr=\mu N=20N$

$fr=m_1{a}_1$

$a_1=5m/s^2$

Now given $m_1$ stops with respect to $m_2$ in 1 sec

hmm..........

u${}_{m_1/m_2}$=-vm/s

v${}_{m_1/m_2}$=0

a${}_{m_1/m_2}$= 5$m/s^2$

t=1 sec

v=u+at

0=-v+5+1

v=5m/s

so m${}_1$was going with speed 5 m/s after m${}_1$ stops with respect to m${}_2$

Let's start fresh

This is a problem of slipping or not slipping.

It's as good as leaving the above setup to motion. Now the question is will m${}_1$ move with m${}_2$ with same acceleration or will m${}_1$ slip on m${}_2$.

Let's choose one and solve

So let's choose all blocks moving together

Oh by the way

When the block m${}_1$ is not slipping on m${}_2$

I can't assume friction to be kinetic (For kinetic friction there has to be relative slipping)

$fr=m_{1}a$

$fr=4a$

20 =10a ............(iii)

a=2$m/s^2$ ........(iv)

well and good but let's not stop here why?

Because we assumed a case, We have to verify whether our solution satisfies our case assumed. I'm sure you do the same in maths.

Okay so assumption was m${}_1$ doesn't slip on m${}_2$ so in that case

fr $\le$ $\mu$${}_k$N

fr $\le$ 20N

we got a = 2 m/s${}^2$

$\Rightarrow$ fr = 8N which satisfies our case.

So our assumption was right.