shubhamsrg
  • shubhamsrg
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
shubhamsrg
  • shubhamsrg
|dw:1356509438621:dw|
anonymous
  • anonymous
concepts of physics?
shubhamsrg
  • shubhamsrg
??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
nthin i thot it was from hc verma well i think i missed smthin fk=5N? why wud they seperate?
shubhamsrg
  • shubhamsrg
pseudo force i'd say..
shubhamsrg
  • shubhamsrg
and i dont know from which book this ques came from,,it was asked in a test..
anonymous
  • anonymous
yup and wat's the acceleration? 10*1
shubhamsrg
  • shubhamsrg
how ?
shubhamsrg
  • shubhamsrg
shouldnt it be 30 -fk = 30 - 5 ?
shubhamsrg
  • shubhamsrg
=25
anonymous
  • anonymous
i dont think fk counts initially they both move as a single body
shubhamsrg
  • shubhamsrg
i see,,i didnt consider that.. anyways, how does it matter? the question will begin only when the block starts moving..
shubhamsrg
  • 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
  • anonymous
opps sorry u r right nd u almost got it
shubhamsrg
  • shubhamsrg
well i'll show you my attempt..just that i didnt get the right ans..
anonymous
  • anonymous
find the independant acc's of both? and find the replative acc and use that goddamn formula
shubhamsrg
  • shubhamsrg
wait, can you confirm to me if the normal reaction acting on the block is 1*10N or (1+2)*10 N ?
anonymous
  • 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
  • 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
  • anonymous
idk...gime sm time i feel giddy
shubhamsrg
  • shubhamsrg
i still find something fishy in my solution.. hmm :|
anonymous
  • anonymous
1 Attachment
shubhamsrg
  • shubhamsrg
how do you say force on block 1 is 10 N? shouldnt the pseudo force be 30 N ?
anonymous
  • anonymous
as far as i kno pseudo is ma
shubhamsrg
  • 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
  • anonymous
ohh u from fiitjee?
shubhamsrg
  • shubhamsrg
no buddy,, :|
shubhamsrg
  • 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
  • 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
  • shubhamsrg
hmmm ..nevermind,,thanks for trying sir! :)
anonymous
  • anonymous
|dw:1356521627848:dw|
anonymous
  • anonymous
So, accleration of the block with respect to the plank = -25/1 = -25ms^2
anonymous
  • anonymous
Now, s=ut+1/2 at^2 2=0+1/2 25 t^2 t^2=100 t=10seconds
shubhamsrg
  • 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
  • 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
  • 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
  • shubhamsrg
just confirm me this, taking into account the pseudo force means A is the relative accn of block right ?
shubhamsrg
  • shubhamsrg
relative to the plank..
anonymous
  • anonymous
Yes, because we are working in the frame of the plank.
shubhamsrg
  • shubhamsrg
i understand this concept a lot better now..thanks..
anonymous
  • anonymous
acceleration w.r.t. ground can be found out as a-A=12.5-7.5=5
anonymous
  • anonymous
.where a and A are the magnitudes. we used - sign because they point in opposite directions.
shubhamsrg
  • shubhamsrg
yep..again,,thanks a ton! :)
anonymous
  • anonymous
you are welcome!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.