 # 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

10 years ago
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
10 years ago

Dear Jeet,

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

25kg×10Nkg1=250N

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

a=250N20kg =12.5ms2

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

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10 years ago

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

10 years ago

2.5 m/s-2

6 years ago
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
6 years ago
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…
4 years ago
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

2 years ago
Dear Student,

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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