Question

A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600

N. What is the tension in the rope if the monkey climbs up with an uniform speed of 5 ms^-1​

Answer 1

★ Correct Question ★

A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What's the tension if monkey climbs up with a uniform speed of $5\:ms^1$

Given :-

• m = 40 kg
• g = 10 m/s
• T_max = 600 N
• S = 5m/s
• a = 0 { Uniform speed i.e, acceleration zero }

Here ,

" m " = Mass of the monkey,

" g " = Acceleration due to gravity.

" T_max " = Maximum tension that the rope can bear.

" a " = Acceleration of the monkey.

" s " = Uniform Speed

Now,

We know, Newton second law of motion equation.

T mg = ma

Applying Newtons second law of motion

From given we know ,

m = 40 kg

g = 10 m/s

T_max = 600 N

a = 0 m/s

Substituting the values in the above equation ,

T mg = ma

Taking " m " common

T = m ( g + a )

↬ 40 (10 + 0)

↬ 40 × 10

↬ 400 N

In this case T < T_max

Hence ,In such case the rope will not break.

_____________________________

Let's assume some more cases for better understanding and to find out in which case the rope will break.

In Case 1

★ If a monkey climbs down with an acceleration of 4 ms −2$.$

Given :-

• m = 40 kg
• g = 10 m/s
• T_max = 600 N
• a = 4 m/s

Here ,

" m " = Mass of the monkey,

" g " = Acceleration due to gravity.

" T_max " = Maximum tension that the rope can bear.

" a " = Acceleration of the monkey.

Now,

We know , Newton second law of motion equation.

T mg = ma

Applying Newtons second law of motion

From given we know ,

• m = 40 kg
• g = 10 m/s
• T_max = 600 N
• a = 4 m/s

Substituting the values in the above equation ,

T mg = ma

Taking " m " common

T = m ( g - a ) { It's in downward }

↬ 40 (10 - 4)

↬ 40 × 6

↬ 240 N

In this case T < T_max

Hence ,In such case the rope will not break.

____________________________

Case 2 :-

★ If he falls down the rope nearly freely under gravity.

Given :-

• m = 40 kg
• In this case monkey falls freely and we know the free fall of object make the acceleration equal to the acceration due to gravity i.e, = 10 m/s
• T_max = 600 N

Here ,

" m " = Mass of the monkey,

" g " = Acceleration due to gravity.

" T_max " = Maximum tension that the rope can bear.

" a " = Acceleration of the monkey.

Now,

T mg = ma

Applying Newtons second law of motion

From given we know ,

m = 40 kg

g = 10 m/s

T_max = 600 N

a = 10 m/s { falling downward }

Substituting the values in the above equation ,

T mg = ma

Taking " m " common

T = m ( g - a )

↬ 40 (10 - 10)

↬ 40 × 0

↬ 0

In this case T < T_max

Hence ,In such case the rope will not break.

_____________________________

Case 3 :-

★ If monkey climbs up with an acceleration of 6 ms -2

Given :-

m = 40 kg

g = 10 m/s

T_max = 600 N

a = 6 m/s

Here ,

" m " = Mass of the monkey,

" g " = Acceleration due to gravity.

" T_max " = Maximum tension that the rope can bear.

" a " = Acceleration of the monkey.

Now,

T mg = ma

Applying Newtons second law of motion

From given we know ,

m = 40 kg

g = 10 m/s

T_max = 600 N

a = 6 m/s

Substituting the values in the above equation ,

T mg = ma

Taking " m " common

T = m ( g + a )

↬ 40 (10 + 6)

↬ 40 × 16

↬ 640 N

In this case T > T_max

Therefore , In such case the rope will break.

____________________________

Answer 2

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Case 1

Substituting the values in the above equation ,

T mg = ma

Taking " m " commonT = m ( g - a )

{ It's in downward }

• 40 (10 - 4)
• 40 × 62
• 40 N

40 NIn this case T < T_max

40 NIn this case T < T_maxHence ,In such case the rope will not break.

Case 2

Now,

Now,T mg = ma

Applying Newtons second law of motion

From given we know ,

m = 40 kg

g = 10 m/s

T_max = 600 N

a = 10 m/s { falling downward }

Substituting the values in the above equation ,

T mg = ma

Taking " m " common

• T = m ( g - a )

• ↬ 40 (10 - 10)

• ↬ 40 × 0

• ↬ 0

In this case T < T_max

Hence ,In such case the rope will not break.