At what minimum acceleration should a monkey slide a rope whose breaking strength is 2/3rd of its weight?
Answers
Answered by
6
Heya........!!!!!!
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Question:-
~~~~~~~~~
At what minimum acceleration should a Monkey slide a rope whose breaking
Strength is2/3 rd of its weight?
Answer
~~~~~~~~
Assume
m be the mass of monkey. So weight = mg />
a be the minimum acceleration of monkey sliding the rope
t be the tension in the rope
By Newton's 3rd law, the net force on monkey can be
F = Weight of monkey — tension in rope
So, F = mg - t where g is the gravitational force.
So the. , tension on an object = (mass of the object x gravitational force)
i.e., t = mg + ma = m(g-a)
(If object is accelerating downwards then the downward force is +ve & upward )
tension is given as 2/3rd of monkey's weight. Sowe can substitute t in the above equation, we get,
(2/3) x mg = m(g - a)
Therefore g=a/3
_____________________________________________________________
Hope it helps
_____________________________________________________________
Question:-
~~~~~~~~~
At what minimum acceleration should a Monkey slide a rope whose breaking
Strength is2/3 rd of its weight?
Answer
~~~~~~~~
Assume
m be the mass of monkey. So weight = mg />
a be the minimum acceleration of monkey sliding the rope
t be the tension in the rope
By Newton's 3rd law, the net force on monkey can be
F = Weight of monkey — tension in rope
So, F = mg - t where g is the gravitational force.
So the. , tension on an object = (mass of the object x gravitational force)
i.e., t = mg + ma = m(g-a)
(If object is accelerating downwards then the downward force is +ve & upward )
tension is given as 2/3rd of monkey's weight. Sowe can substitute t in the above equation, we get,
(2/3) x mg = m(g - a)
Therefore g=a/3
_____________________________________________________________
Hope it helps
Answered by
0
Let a be the acceleration of monkey.
The apparent weight =mg−ma=2mg/3
or ma=mg(1−2/3)=mg/3⇒a=g/3
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