IIT JEE , AIEEE Forum - Physics , Mechanics - Challenge

karthik2007

Try this one out. I won't say that it is tough. Just requires your basics to be good. In fact, it is somewhat easy I would say.

A block of mass m is pulled with a constant force of magnitude mg/2 along a horizontal surface (assume it to be smooth). The force makes an angle

with the horizontal which varies as

= as where s is the displacement of the block and a is some constant. Find the velocity of the block, as a function of the angle at any instant that the force makes with the block.

rooney

Let theta be k. Displacement be s.
So, k=as
dk=a ds

Let force at any moment of time t make angle k
Then at time t + dt, it will make angle k +dk

So, at t+dt,
mgCos(k + dk)/2 = ma
= mvdv/ds
gCos(k+dk)/2 = avdv/dk

Cos(k+dk) can be taken as equal to Cosk

Integrating,
g(Sink)/a= v² where k varies from 0 to theta

v= (gsin

/a)^1/2

karthik2007

Any other attempts? I'll tell which answer is right tomorrow

waterdemon

Hello Karthik,
Here is my solution.

I have given the free Body diagram along with my solution.

FSin@ + R = mg.............(1)
FCos@ = ma₀ .............(2)

Now from Equation (2).

(mg/2)Cos(as) = mx(dv/dt)
(mg/2)Cos(as) = mv(dv/ds)

in the above statement multiplying and dividing by v=ds/dt

_[0]

^[v] v.dv = g/2

Cos(as).ds

From solving above we will get:
v²/2 = (g/2a)Sin(as) + C

Now when v=0 , s=0 , C=0
Therefore,
v² = (g/a)Sin(as)

And hence,
v =

gSin(as)
a

Hope you find it useful.
Rate if useful.

Cheers!!!!!!!!!!!!!!!

karthik2007

Yes, both the solutions are right :)

waterdemon

Ya dude,
Both are right but I have given a bit different method.

karthik2007

Why did you write the step Fsin@ + R = mg? That equation is never really considered when it comes to calculations.So why that equation?

waterdemon

Hey Dude,
Don't you know the first law of solving.
START FROM BASIC STEPS SLOWLY AND STEADILY.

I don't solve it on paper.
I solved i while typing.

I thought that may be the equation gets in use but no it
was of no use so is that my fault.

Cheers!!!!!!!!!!@@@!!!!!!!!!

Anyways I don't want to make anymore comments on this topic.
Enjoy!!!!!!!!

karthik2007

My basics are not as hard core as yours yaar.. I am just an ordinary boy.. but here the velocity is asked, so isn't it obvious that we don't have to consider the vertical displacement? (Block is going to move on the floor only yaar)

Anyway, your solution is good (as always.. excellent presentation :) ). Looks good too with the diagram :)

Cheero!

shubham_sachdeva

well i have given a try & got a bit different answer..

as plane is smooth so friction absent.....hence no heat dissipated...

so applying energy conservation we get

1/2 m v^2 = work done by force F.(F = mg/2)

1/2 mv^2 = (mg/2)(s)(cosk)

solving we get

v = (gscosk)^1/2

also, k = as , so s = k/a putting above we get

v = (gk/a.cosk)^1/2 -------ans..

plzz point out my mistake...

karthik2007

@Shubam - you can use the equation work done = F.S only if the force component F is a constant, which is not in this case.

rajat.khanduja

Pls point out my mistake as i've used a dif. method.

Along X-axis

m(v_x)² /2 =

F cos(as) ds
= mg/(2a) sin(as)

hence, (v_x)² = g/a * sin(as)

Similarly

(v_y)² = (-g/a) * cos(as)

Therefore v =

[ (v_x)²+ (v_y)²]

=

g/a { sin(as) - cos (as)

karthik2007

The block moves only in the horizontal direction

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