IIT JEE , AIEEE Forum - Physics , Mechanics - Good Questions from Work Energy Power..Try them

goiit_user

Q.1
A ring of mass m slides on a smooth vertical rod. Attached to the ring is a light string passing over a smooth peg at a distance b from the rod, and at the other end of string is a mass M (>m). The ring is held on a level with the peg and released.
Show that it first comes to rest after falling a distance :
(2mMa)/ (M²- m²)...
See Fig.1

Q.2
A small body A starts sliding from the height h down an inclined groove passing into a half-circle of radius h/2. Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory( after breaking off the groove.)
Ans = (2/3)*

(gh/3)
See Fig.2

Q.3
A particle is projected along the inside of a smooth fixed sphere, from the lowest point, with a velocity equal to that due to falling freely down the vertical diameter of the sphere. Show that the particle will leave the sphere and afterwards pass vertically over the point of projection at a distance equal to 25/32 of the diameter.

Q.4
A thin rim of mass m and radius r rolls down an inclined plane of slope

, winding thereby a ribbon of linear density

. At the initial moment, the rim is at a height h above the horizontal surface.
Determine the distance S from the foot of the inclined plane at which the rim stops, assuming that the incline plane smoothly changes into the horizontal plane.
Ans : S = (mg +

(hcosec

)(r - h/2) /

r
See Fig. 3

Q.5
A chain AB of length L is loaded in a smooth horizontal tube so that its fraction of length h hangs freely and touches the surface of the table with its end B. At a certain moment, the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube.
Ans :

(2gh log_e(l/h))
See Fig. 4

Q.6
A body of mass m was slowly hauled up the hill by a force E which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l and coefficient of kinetic fraction k.
Ans : mg (h+ kl)
See Fig.5

goiit_user

Q.7
A small bar resting on a smooth horizontal plane is attached by threads to a point P and by means of weightless pulley, to a weight B possessing the same mass as the bar itself. Besides, the bar is also attached to a point O by means of a light non-deformed spring of lenght l₀ = 50cm and the stiffness K = 5mg/l₀, where m is the mass of the bar. The thread PA having been burned, the bar starts moving. Find its velocity at the moment when it is breaking off the plane.
Ans : 1.7 ms^-1
See Fig. 6

Q.8
A horizontal plane supports a plank with a bar of mass m=1 kg placed on it and attached by a light elastic non-deformed cord of length l₀ = 40 cm to a point O. The coefficient of friction between the bar and the plank equals k=0.20.
The plank is slowly shifted to the right until the bar starts sliding over it. It occurs at the moment when the cord deviates from the vertical by an angle 30^o. Find the work that has been performed at that moment by the frictional force acting on the bar in the reference frame fixed to the plane.
Ans : W = 0.09 Joule
See Fig. 7

karthik2007

I am getting the velocity for the second one as sqrt(gh/3)...

could you recheck? Is the breaking angle cos@ = 2/3 with the vertical?

goiit_user

Some one told me that we need to find vcos

..and not v...
So ur answer is correct.
Plz provide explanation of how u got breaking angle and v

karthik2007

Okay... just follow the soln carefully... better you draw an FBD.

Now, let the body make an angle @ withe the vertical at the instant in breaks off...

So conserving energy, we get:

mgh = 1/2mv² + mgh(1+cos@)/2

Solving for v², we get v² = gh(1-cos@)

Also, balancing forces on the body, we get:

mv²/(h/2) = mgcos@ + N..

now N = 0,

so from this, we get v² = ghcos@/2

Equating the two expressions for v² we get cos@ = 2/3.

So, v = sqrt(gh/3)

But I don't think u need to find vcos@ here... doesn't make sense... did u check the book ans? which book is this from?

karthik2007

Q6.

using Work Energy theorem, we get:

Work done by E = change in PE + wd by friction..

Hence, we get WD = mgh + kmgl

or WD = mg(h+kl)

goiit_user

I think the reason for vcos

, as I think, is that the body will go through a parabolic path once it leaves the groove..and at the highest point the only velocity that exists will be that in the x-direction.. That's why we take vcos

and not v..

karthik2007

For the first question... I am not getting that darn 2!!!

goiit_user

I've tried all these questions a lot but cud not come up with the answer!!

Tell me how have u approached for the 1st question..maybe I can get that 2 also!!

karthik2007

I know how to do that chain question... simple integration only.

As for the first question... I am not sure if I've balanced the forces correctly. Let the string make an angle of @ with the horizontal... now find Tension... get a relation between @ and the masses... then use trigonometry to get the ans. This is the approach

karthik2007

btw Q6 is also solved... look above in case u didnt see

waterdemon

Arey karthik you have done everything right for the 2nd
question but just one thing thats not getting to ur mind.

See:
After breaking off the groove,the body will move like a
projectile forming an angle @ with the horizontal.At the
Highest point of the Trajectory, the vertical component
of the velocity will be Zero.

Hence,
V_f = VCos@
V_f =

gh/3 * (2/3)

Hence V = (2/3)

gh/3

Hope it is useful.
Rate if useful.

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

karthik2007

Excellent waterdemon...!!! Now it has entered my mind clearly :)

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