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It appears (though not clearly mentioned) that 40 N force is perpendicular & horizontal against the wall.
max frictional force = 0.5*40 = 20 N (in a direction opposite the movement of the the mass)
Forces acting on the mass is -
(i) its weight (2*g=19.6 N) acting vertically downwards parallel to the wall)
(ii) 15 N horizontal & parallel to the wall
The mass can move only parallel to the wall.
The frictional force will act opposite the resultant of the above two forces
The resultant of the above two forces is (225+19.6**0.5)**0.5 which is >20
Hence the block will move
the direction of the movement will tan inverse 15/19.6 from the vertical downwards
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I see no reason for negative answer as the question asked is the work done on the mass by the external field. which is F.dr and
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In the problem no.1 of max ration of h/b, i think the condition for maximum h/b is to found out is to be given.
The condition may be like max h/b for which the P>0 and the block just begins to move...
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yes answer is mg/2.. Thanks savvej..good explanation!
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| Q.A block of mass m is placed on a trolley free to move on smooth horizontal surface. What maximum force can be applied on trolley so that block doesn't slide? Mass of troll is 2 m and co-efficient of friction between block and trolley is = 0.5 |  | | A | 3 mg | B | 3/2 mg | | B | mg | A | mg/2 | |
(This is same problem as being shown on the home page of GOIIT.com , the figure accompanying the problem unable to be shown..It can be taken from home page of GOIIT.com
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look, at 50 mtrs height vertical component of initial velocity becomes zero.
So, vertical component of initial velocity = [ ] 2gh = [ ] 2*10*50 (g= 10 m/s^2)
vertical component of initial velocity = 10[ ] 10 m/s
time taken to reach 50 mtrs high = 10[ ] 10 /g =[ ] 10 sec
horizontal component of initial velocity = 100/[ ] 10 [window is 100 mtr away]
horizontal component of initial velocity = 10[ ] 10 m/s
total velocity = [ ] (1000+1000) = 44 m/s approx.) & since both the component are equal , the direction is 45 degrees
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just observe that
in case of (b) mass 5 kg will come down with half the rightward acceleration of mass 2 kg AND the tension in rope is same everywhere.
Similarly , in (c), the acceleration of mass 1 kg (upwards) will be half the downward acceleration of mass 2 kg AND tension in the rope will be same everywhere
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IIn my opinion, the heat produced will be same as the relative motion shouldn't have any effect on heat produced and measured.
But , the kinetic energy loss/gain in both the cases will be different.
The reason is simple .
say a particle is moving with velocity V relatice ground at rest. Its kinetic Energy is 1/2 mV^2.
Now , if an observer is also moving with velocity V, the relative velocity is zero and KE observed by this observer is zero.
I think this is what led Einstein to derive his theory of relativity.
Too bad Einstein was born a little earlier....who knows an Indian would have been hailed as .......!!!!!!!!!!!!! Anyway..such is life...
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interesting question....
you apply a force on the wall...as wall is fixed to the ground there is an equal and opposite reaction o the wall from the ground...so no acceleration...
pl use normal englis!
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answer seems to be 121/(6x6x6x6)
is it correct?
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equate the kinetic energy of the moving block to 1/2.kx^2 and find x. this is the maximum displacement
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why two displacements x1 & x2. It has to be same x. now consider mass m1 upon whom F1 ia acting. let's say the accelaratio of the system is a (= [F1-F2]/[m1+m2]) F1-kx=m1.a Kx=F1-m1a=F1-m1(F1-F2)/(m1+m2) x= 1/k[(m2.F1+m1.F2)/ (m1+m2)] you can check the correctness of the answer by putting m1=m2 and/or F1=F2 when F1=F2,there will be no acceleration and extention x= F/k , simple. If m1=m2 , x= (F1+F2)/2k, again simple.
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Dear Dunnol,
When a vehicle is 'turning ' , its wheels have turned in that direction though rest of the body continues to move with straight velocity. The wheels roll. THey pull the whole car in their 'turned direction', but the rest of the body has its momentum in the tangential direction (that is, if turning is circular). So the body of the car has to be 'pulled' by the wheels in a different direction' This requires a force on the tyres which is radially inward (you can visualise that).
Hope it helps!
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dear ashish , your question is ok except it requires coefficient of friction to be given...it'll start rolling (pure)when frictional torque is equal to weight-torque.
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this will give rolling with initial velocity! The question should be : On a frictional surface, the disc with radius R is given an initial linear velocity v. At what speed the disc will start pure rolling. [Note , in this case initially there will be slipping and rolling , while in case if the disc is rolled initially, ity will roll from the beginning itself..]
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it needs to be clarified the probability of c supporting A or B and whether solution offered by {A,B,C} needs to be supported by all three .
if C rarely supports A or B , then how this team offers solution
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there is no reason why coefficient of friction has to be less than 1.
But spare a thought... GOD is a fantastic engineer...HE can't design something whic has problem in runing...coefficient of friction >1 will mean that you need to apply greater force to push a thing than to lift it...it is almost like an attractice force acting at 90 deg to the line of bodies' surface...something which appears to be little odd, though it doesn't violate any physical laws known so far....!!!!
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dear yaz....i like your approach..
but gases can chemically react...and expand...molecular expansions...one molecule providing a surface to react to another...
it's more like shooting of the gun (have u ever participated in yr NCC gun-shooting practice..?) it's same...Gun cartridge expands chemically and runs away in forward direction and pushes the barrel of the gun in reverse direction...No surface neede here...
Similarly fuel expands pushing air ship in opposite direction..
Another example is of balloon....if you untie a expanded baloon...it moves in opposite direction...No surface needed...
hope it helps
pl do rate me
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since block B is described as small and no mass is given...assume a neglible mass for it..
there fore he acceleration produced in the block A =10/5=2 m/s^2
block B will have a relative acceleration of 2 m/s^2 ...time taken to cover a distanmce of 0.20 m is [ ] 0.4/2 =[ ] 0.2 sec
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I also join the fray.......!!!!!!!!!!!!!!!!!!
How can answer not be 11 Kg...
No need to invoke Archmides...
Simple..total weight of the things put on the balance is 11 kg ...we are measuring the weights OUTSIDE the water....Archimedes gets into action only when you are weighing something under water REMEMBER that.
Also just imagine...if you keep the block on the balance but outside the water...the balance will weigh 11 kg....how can it be any different if you put the block inside the water...as far as balance is concerened , both the situations are same...
hope it helps
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