a block of mass 1kg is placed on a plank of mass 2kg, length of plank = 2m. Coefficient of friction between plank and block is 0.5 and the round on which the plank is kept is smooth.
A constant force F=30N is applied on the plank; the time after which the block separates from the plank =?

- shubhamsrg

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- shubhamsrg

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- anonymous

concepts of physics?

- shubhamsrg

??

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## More answers

- anonymous

nthin i thot it was from hc verma
well
i think i missed smthin fk=5N?
why wud they seperate?

- shubhamsrg

pseudo force i'd say..

- shubhamsrg

and i dont know from which book this ques came from,,it was asked in a test..

- anonymous

yup and wat's the acceleration?
10*1

- shubhamsrg

how ?

- shubhamsrg

shouldnt it be
30 -fk = 30 - 5 ?

- shubhamsrg

=25

- anonymous

i dont think fk counts initially
they both move as a single body

- shubhamsrg

i see,,i didnt consider that..
anyways, how does it matter? the question will begin only when the block starts moving..

- shubhamsrg

i mean the solution will begin*
also, fk isnt high enough relative to the force, so shouldnt it start moving at t=0 only ?

- anonymous

opps sorry u r right
nd u almost got it

- shubhamsrg

well i'll show you my attempt..just that i didnt get the right ans..

- anonymous

find the independant acc's of both?
and find the replative acc
and use that goddamn formula

- shubhamsrg

wait, can you confirm to me if the normal reaction acting on the block is 1*10N or (1+2)*10 N ?

- anonymous

wat's block 1?
anyways i think i understand ur confusion
well i guess it is 10 but
u jus need to find fk....and draw fbd with the horizontal forces

- shubhamsrg

allow me to try
accn of block w.r.t to ground = 30 -5 /1 = 25
accn of plank w.r.t to ground = 30 - 5 /2 = 12.5
relative accn = 37.5
is that right?

- anonymous

idk...gime sm time i feel giddy

- shubhamsrg

i still find something fishy in my solution.. hmm :|

- anonymous

##### 1 Attachment

- shubhamsrg

how do you say force on block 1 is 10 N?
shouldnt the pseudo force be 30 N ?

- anonymous

as far as i kno pseudo is ma

- shubhamsrg

well,,i'd argue there,,i maybe wrong,,
anyways, that still doesnt yield the right ans, so got to be some flaw there..

- anonymous

ohh
u from fiitjee?

- shubhamsrg

no buddy,, :|

- shubhamsrg

am dropping my year for jee and ISI and CMI etc, living here in a remote place, just a small tuition like coaching here, no fitjee,resonance,akash here ,nearby i mean..

- anonymous

ohh i get it....it's okay......u have a better chance than any1 in here
nd ur question....sry it's been long since i worked on these.........so i donno if i can help

- shubhamsrg

hmmm ..nevermind,,thanks for trying sir! :)

- anonymous

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- anonymous

So, accleration of the block with respect to the plank = -25/1 = -25ms^2

- anonymous

Now,
s=ut+1/2 at^2
2=0+1/2 25 t^2
t^2=100
t=10seconds

- shubhamsrg

@sauravshakya you missed out on the fact that the plank also moves, so the block wont have to cover 2m completely, if you get what i mean.

- anonymous

When the block starts to move the maximum friction acts on it i.e.5N
Then we draw the free body diagram as shown in the file attached. Note that we work in the frame of the plank because it is easier. Pseudo force is given by "ma". Where m is the mass of the body in concern(i.e. in this case 1kg) and "a" is the acceleration of the frame(i.e. the plank). We get the free body diagrams for plank and block as shown.
Eqn of motion for plank is 30-5=2a giving a=12.5m/s/s
Eqn of motion for block is 1*a-5=1*A where A is the acc of the 1kg block
Inserting the value of "a" gives A=7.5m/s/s
Length of plank is 2m
Using 2nd eqn of motion we get time to cover 2m =0.73s(approx)
***DO TELL THE CORRECT ANSWER PLEASE**

##### 1 Attachment

- shubhamsrg

thats where i went wrong, so it was a conceptual problem from my side..thanks a lot @Diwakar
yep, thats the correct ans

- shubhamsrg

just confirm me this, taking into account the pseudo force means A is the relative accn of block right ?

- shubhamsrg

relative to the plank..

- anonymous

Yes, because we are working in the frame of the plank.

- shubhamsrg

i understand this concept a lot better now..thanks..

- anonymous

acceleration w.r.t. ground can be found out as a-A=12.5-7.5=5

- anonymous

.where a and A are the magnitudes. we used - sign because they point in opposite directions.

- shubhamsrg

yep..again,,thanks a ton! :)

- anonymous

you are welcome!!

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