Question

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and reducing the force on the climber. In one standardized test of ropes, an 80 kg mass falls 4.8 m before being caught by a 2.5-m-long rope. If the net force on the mass must be kept below 11 kN, what is the minimum time for the mass to come to rest at the end of the fall

Answer 1

The minimum time for the mass to come to rest at the end of the fall is about 0.071 s

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

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$\large {\boxed {F = \Delta (mv) \div t }$

F = Force ( Newton )

m = Object's Mass ( kg )

v = Velocity of Object ( m/s )

t = Time Taken ( s )

Let us now tackle the problem !

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Given:

mass of climber = m = 80 kg

height of fall = h = 4.8 m

net force = ∑F = 11 k N = 11000 N

Asked:

minimum time = t = ?

Solution:

FIrstly , we could find the initial velocity of climber as he caught by the rope:

$v^2 = u^2 + 2gh$

$v^2 = 0^2 + 2(9.8)(4.8)$

$v^2 = 94.08$

$v = \frac{28}{3}\sqrt{5} \texttt{ m/s}$

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Next , we will use Newton's Law of Motion to calculate the minimum time:

$\Sigma F = \Delta p \div t$

$\Sigma F = m \Delta v \div t$

$11000 = 80 (\frac{28}{3}\sqrt{5}) \div t$

$t = 80 (\frac{28}{3}\sqrt{5}) \div 11000$

$t \approx 0.071 \texttt{ s}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer 2

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$Final velocity

$v_i =$Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s