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Question

In the shown diagram m1 = m2 = 4 kg and m3 = 2 kg.Coefficient of friction between m1 and m2 is 0.5. The mass m1 is given a velocity v and it just stops at the other end of the mass m2 in 2 sec. Let a1, a2 and a3 be the acceleration m1, m2 and m3 respectively, then:


A

For t<1sec,,

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B

For t<1sec,,

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C

the value of v is 5 m/s

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D

For t<1sec,

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Solution

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.

m3 will move down. m2 will slide rightwards m3 and m2 are connected by inextensible massless string so acceleration of these 2 are going to be same for sure

a2 = a3 = a

Given m1 is sliding on m2 in leftward direction with respect to m2

So friction on m1 will be backward or is rightward direction and according to 3rd law, equal force will be on m2 in leftward direction.

Let's draw free body diagram:

m3gT=m3a

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

T - fr=m2a

fr=μN=μm1g=12×40=20N

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

From (I) & (II) a2 = a3 = a = 0

So m2 & m3 won't move till the block m1 is in relative motion with respect to m2

N=m1g

fr=μN=20N

fr=m1a1

a1=5m/s2

Now given m1 stops with respect to m2 in 1 sec

hmm..........

um1/m2=-vm/s

vm1/m2=0

am1/m2= 5m/s2

t=1 sec

v=u+at

0=-v+5+1

v=5m/s

so m1was going with speed 5 m/s after m1 stops with respect to m2

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 m1 move with m2 with same acceleration or will m1 slip on m2.

Let's choose one and solve

So let's choose all blocks moving together

Oh by the way

When the block m1 is not slipping on m2

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

fr=m1a

fr=4a

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

a=2m/s2 ........(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 m1 doesn't slip on m2 so in that case

fr μkN

fr 20N

we got a = 2 m/s2

fr = 8N which satisfies our case.

So our assumption was right.


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