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
B
m=Wgga
C
m=W(1+ag)
D
the tension in the rope is Wg N

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 $$m=\dfrac{Wg}{(g-a)}$$Physics

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