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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 =?

Physics
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concepts of physics?
??

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Other answers:

nthin i thot it was from hc verma well i think i missed smthin fk=5N? why wud they seperate?
pseudo force i'd say..
and i dont know from which book this ques came from,,it was asked in a test..
yup and wat's the acceleration? 10*1
how ?
shouldnt it be 30 -fk = 30 - 5 ?
=25
i dont think fk counts initially they both move as a single body
i see,,i didnt consider that.. anyways, how does it matter? the question will begin only when the block starts moving..
i mean the solution will begin* also, fk isnt high enough relative to the force, so shouldnt it start moving at t=0 only ?
opps sorry u r right nd u almost got it
well i'll show you my attempt..just that i didnt get the right ans..
find the independant acc's of both? and find the replative acc and use that goddamn formula
wait, can you confirm to me if the normal reaction acting on the block is 1*10N or (1+2)*10 N ?
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
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?
idk...gime sm time i feel giddy
i still find something fishy in my solution.. hmm :|
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how do you say force on block 1 is 10 N? shouldnt the pseudo force be 30 N ?
as far as i kno pseudo is ma
well,,i'd argue there,,i maybe wrong,, anyways, that still doesnt yield the right ans, so got to be some flaw there..
ohh u from fiitjee?
no buddy,, :|
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..
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
hmmm ..nevermind,,thanks for trying sir! :)
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So, accleration of the block with respect to the plank = -25/1 = -25ms^2
Now, s=ut+1/2 at^2 2=0+1/2 25 t^2 t^2=100 t=10seconds
@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.
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**
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thats where i went wrong, so it was a conceptual problem from my side..thanks a lot @Diwakar yep, thats the correct ans
just confirm me this, taking into account the pseudo force means A is the relative accn of block right ?
relative to the plank..
Yes, because we are working in the frame of the plank.
i understand this concept a lot better now..thanks..
acceleration w.r.t. ground can be found out as a-A=12.5-7.5=5
.where a and A are the magnitudes. we used - sign because they point in opposite directions.
yep..again,,thanks a ton! :)
you are welcome!!

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