Answer :
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
[tex]F = ma[/tex]
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{11*10^3}{80}[/tex]
[tex]a = 137.5m/s[/tex]
From the cinematic equations of motion we know that
[tex]v_f^2-v_i^2 = 2ax[/tex]
Where,
[tex]v_f =[/tex]Final velocity
[tex]v_i =[/tex]Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
[tex]v_f^2-v_i^2 = 2ax[/tex]
[tex]0-v_i^2 = 2(-g)x[/tex]
[tex]v_i =\sqrt{2gx}[/tex]
[tex]v_i = \sqrt{2*9.8*4.8}[/tex]
[tex]v_i = 9.69m/s[/tex]
From the equation of motion where acceleration is equal to the velocity in function of time we have
[tex]a = \frac{v_i}{t}[/tex]
[tex]t = \frac{v_i}{a}[/tex]
[tex]t =\frac{9.69}{137.5}[/tex]
[tex]t = 0.0705s[/tex]
Therefore the time required is 0.0705s
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.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
[tex]\texttt{ }[/tex]
[tex]\large {\boxed {F = \Delta (mv) \div t }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
v = Velocity of Object ( m/s )
t = Time Taken ( s )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
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:
[tex]v^2 = u^2 + 2gh[/tex]
[tex]v^2 = 0^2 + 2(9.8)(4.8)[/tex]
[tex]v^2 = 94.08[/tex]
[tex]v = \frac{28}{3}\sqrt{5} \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
Next , we will use Newton's Law of Motion to calculate the minimum time:
[tex]\Sigma F = \Delta p \div t[/tex]
[tex]\Sigma F = m \Delta v \div t[/tex]
[tex]11000 = 80 (\frac{28}{3}\sqrt{5}) \div t[/tex]
[tex]t = 80 (\frac{28}{3}\sqrt{5}) \div 11000[/tex]
[tex]t \approx 0.071 \texttt{ s}[/tex]
[tex]\texttt{ }[/tex]
Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
- Newton's Law of Motion: https://brainly.com/question/10431582
- Example of Newton's Law: https://brainly.com/question/498822
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Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics
