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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............................................................

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............................................................

Grade:12

2 Answers

AYUSH SHARMA
33 Points
10 years ago
hiiiiiiiiiiiiiiiiiiiiii dude please approve this
vikas askiitian expert
509 Points
10 years ago

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

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