__Question 5.33__:(a) climbs up with an acceleration of 6 m s

^{–2}

(b) climbs down with an acceleration of 4 m s

^{–2}

(c) climbs up with a uniform speed of 5 m s

^{–1}

(d) falls down the rope nearly freely under gravity?

*(Ignore the mass of the rope).*

__Solution__:

**Case (a)***m*= 40 kg

Acceleration due to gravity,

*g*= 10 m/s

Maximum tension that the rope can bear,

*T*

_{max}= 600 N

Acceleration of the monkey,

*a*= 6 m/s

^{2}upward

Using Newton’s second law of motion, we can write the equation of motion as:

*T*–

*m*g

*= ma*

∴

*T*=

*m*(g

*+ a*)

= 40 (10 + 6)

= 640 N

Since

*T*>

*T*

_{max}, the rope will break in this case.

**Case (b)**Acceleration of the monkey,

*a*= 4 m/s

^{2}downward

Using Newton’s second law of motion, we can write the equation of motion as:

*m*g –

*T*=

*ma*

∴

*T*=

*m*(g

*– a*)

= 40(10 – 4)

= 240 N

Since

*T*<

*T*

_{max}, the rope will not break in this case.

**Case (c)**The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e.,

*a*= 0.

Using Newton’s second law of motion, we can write the equation of motion as:

*T*–

*m*g

*= ma*

*T*–

*m*g = 0

∴

*T*=

*m*g

= 40 × 10

= 400 N

Since

*T*<

*T*

_{max}, the rope will not break in this case.

**Case (d)**When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e.,

*a =*g

Using Newton’s second law of motion, we can write the equation of motion as:

*m*g –

*T*=

*m*g

∴

*T*=

*m*(g

*–*g) = 0

Since

*T*<

*T*

_{max}, the rope will not break in this case.

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