12.
tension is F.
20 J is the sum of the work done by F and mg
1)Consider two observers moving with respect to each other at a speed v along a straight line. they observe a block of mass m moving a distance l on a rough surface. Which of the following quantities will be same as observed by the two observers and why????
a)kinetic energy of the block at time t
b)work done my friction
c)total work done on the block
d)acceleration of the block
2)A block of mas M is hanging over a smooth & light pulley though a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1 second, then
a) the tension in the string is Mg
b)the tension in the string is F
c)The work done by the tension on the block is 20 J
d) the work done by the force of gravity is -20 J
3) A particle of mass m is attached to a light string of length l, the other end of which is fixed, Initially the string is kept horizontal and the particle is given an upward velocity v, the particle is just able to complete a circle.
a) The string becomes slack when the particle reaches its highest point
b)the particle again passes through the initial position.
why option (a) is correct ????????
4)A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is (u). Find the work done by the friction during the period the chain slips off the table.
5)A smooth of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at 0 velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as function of the angle theta it slides.
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11 Answers
491. d)acceleration of the block
and that is because the 2 observers are NOT acceleratng w.r.t. each other!
11. b) work done by friction also
as there is no relative motion at contact surfaces
11Thanks Subhomoy your answer is correct
thanks Anirudh
Thanks Nithin, but your answer is incorrect
493) when the body "just" completes a circle, we ensure that on top, v=0 and T=0
Thus the only force acting is mg downwards....
obviously the string slacks!! :)
11If the string becomes slack, so how will the particle completes the circle.?
And when the particle is just able to complete the circle its velocity at the highest point is √gR and the Tension = 0
49oops...i just messed up everything!!
actually the vel is not 0...
the tension is 0
and thats because in the frame of reference of the rotating block,
weight is balanced by centrifugal force again i remind as i always do..do not chuckle at this word! :P
and so T=0
thus string slacks..
and though the string slacks, due to horizontal velocity, the mass moves on and completes the circle1!
14) ans is -2uMgL/9
Use the fact that the friction force does work to shift the COM of the part of chain that lies on the table to the edge...
23since it is an extreme case, the minimum possible force on the particle in the radial direction shud be able to provide the exact reqd centripetal force.