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Question

A monkey of mass $$m\ kg$$ slides down a light rope attached to a fixed spring balance, with an acceleration $$a$$. The reading of the spring balance is W kg. [g $$=$$ acceleration due to gravity]. Then:


A
the force of friction exerted by the rope on the monkey is m(ga) N
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B
m=Wgga
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C
m=W(1+ag)
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D
the tension in the rope is Wg N
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Solution

The correct options are
A the force of friction exerted by the rope on the monkey is $$m(g - a)\ N$$
B the tension in the rope is $$Wg\ N$$
D $$\displaystyle m = \dfrac{Wg}{g - a}$$
Let the force of friction between the rope and the monkey is $$f$$. 
now $$ma=mg-f \Rightarrow f=m(g-a) N$$
Here , the reading in spring balance $$=$$ tension in the rope.
so the tension in the rope is $$T=Wg N$$
also, 
the force of friction between the rope and the monkey $$=$$ tension in the rope.
so, $$m(g-a)=Wg$$
or $$m=\dfrac{Wg}{(g-a)}$$
173586_138220_ans_1bb7b90653fe44c786e9caf9ff7e8cdd.png

Physics

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