Question

A monkey of mass $mkg$ 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 $m(g-a)N$
• B. $m=\frac{Wg}{g-a}$
• C. $m=W(1+\frac{a}{g})$
• D. the tension in the rope is $WgN$

Answer 1

Answer: (A), (B), (D)

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 $WgN$
D $m=\frac{Wg}{g-a}$

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 $T=WgN$
also, the force of friction between the rope and the monkey $=$ tension in the rope.
so, $m(g-a)=Wg$
or $m=\frac{Wg}{(g-a)}$