Question

7. A monkey is climbing on the rope as shown in
figure. The string and the pulley are light. The
acceleration of monkey relative to rope is 2m/s?
downwards. The acceleration of the monkey is
1 - 11/2 m/s^2
2- 6m/s^2
​

Answer 1

Answer:

Suppose the monkey acceleration upward with acceleration 'a' & the block, acceleration downward with acceleration a

1

. Let Force exerted by monkey is equal to T

From the free body diagram of monkey∴T−mg−ma=0

⇒T=mg+ma.Again from the FBD of the block

T=ma

11

−mg=0

⇒T=mg+ma+ma

1

−mg=0 [From (I)]

⇒ma=−ma

1

⇒a=a

1

Acceleration −a downward i.e. a upward.

∴ The block & the monkey move in the same direction with equal acceleratio.If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they have same weight (same mass). Their separation will not changed as time passes.

Answer 2

A monkey is climbing on the rope as shown in figure. The string and the pulley are light. The acceleration of monkey relative to rope is 2m/s² downwards.

The acceleration of the monkey is ...

see the diagram attached in answer,

let the acceleration of string is a and the acceleration of monkey is A.

for mass, m = 5 kg ,

tension in the string - weight of mass = ma

⇒ T - 5g = 5a ....(1)

for monkey of mass , M = 15 kg ,

weight of monkey - tension in the string = MA

⇒ 15g - T = 15A ...(2)

from equations (1) and (2),

⇒ 10g = 5a + 15A ....(3)

as it has given that the acceleration of monkey relative to rope is 2 m/s².

∴ A - a =2 ...(4)

from equations (3) and (4), we get, A = 11/2

therefore the acceleration of monkey is `11/2 m/s²