A monkey of mass 20 Kg is holding a vertical rope. The rope can break when a mass of 25 Kg is suspendedfrom it. What is the maximum acceleration with which the monkey can climb up along the rope?(a)2.5 ms-2 (b) 5 ms-2 (c) 7 ms-2 (d) 10 ms-2

Let the monkey climb up with acceleration a. Then tension caused in the rope will be T =

M(g+a), where M is the mass of the monkey. Maximum value of T is 25g = 20(g + a). Taking

g = 10ms

-2, it gives a = 2.5 ms

Approved

Dear Jeet,

assuming the value of g is 10ms^-2, the maximum weight that the rope can hold is  

. (though it should be less than 250, I''m not so sure about the maimum part) using F=ma,

The maximum acceleration of the monkey is 12.5ms^-2

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tension is responsiblefor upward acc. and  maximum tension it can bear is 250N. hence max acc.

250-200=20*a

a=2.5m/s^2

2.5 m/s-2

A monkey of mass 20 kg. is holding a vertical rope. The rope will not break when a mass of 25 kg. is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration it which the monkey can climb up along the rope

let the initial tension be Tusing fbd for m=20kg we get,T-mg=maT-20g=20a…1Tmax-25g=0Tmax=25g……2from 1,225g-20g=20a5g=20aa=g/4 =2.5m/s^2hope this helps…

Dear Student,
Please find below the solution to your problem.

1kgf = 1kg * 10m/s square
=10N
For 25kgf=25kg * 10m/s square
=250N _____(1)
Monkey mass =20kg so mg=200N (2)
Subtract equation 2 from 1 we get
250N-200N=50N
Force = Mass *acceleration
50 =20kg * a
a = 2.5 m/s square

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