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rajan jha Grade: 12

A monkey pulls the midpoint of a 10cm long light inextnsible string connecting two identical objects A and B lying on the smooth table of masses 0.3kg continously along the perpendicular bisector of line joining the masses. the masses are found to approach each other at a relative acceleratioin of 5m/s2 when they are 6cm apart. the constsnt force applied by monkey is............................................................

7 years ago

Answers : (2)

33 Points
										hiiiiiiiiiiiiiiiiiiiiii dude please approve this
7 years ago
vikas askiitian expert
510 Points

let length of string is 2l

monkey is pulling from mid point , let after some time its particles are at 2x distance from each other then

there will be similar  triangles whose hypotenuse is l & one of its side is x....

other side will be (l2-x2)1/2

2Tcos@ = F      (force applied by monkey)

 T = F/2cos@

 ma = F/2cos@

 a = F/2mcos@............1

from figure cos@ = (l2-x2)/l               (@ is angle bw perpendicular bisector & tension acting on any of particle)

a = Fl/2m(l2-x2)

2l = 10 , l =5cm

2x = 6 ,x =3

putting these values

a = 5F/8m

5 = 5F/8m                      (a = 5m/s2 (given))

F = 8m = 2.4N

7 years ago
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