Answer:
0.967 m/s
1855 N
[tex]46.375\ \text{MPa}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement = 4.77 cm
g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]
The velocity of the object at the moment of impact is 0.967 m/s
Now
[tex]\Delta v[/tex] = Change in velocity = 1 m/s
t = Time taken = 20 ms
m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]
[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]
The force associated with a single step of the person is 1855 N
A = Area = [tex]0.4\ \text{cm}^2[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]
The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]
The speed of object during falling is 0.967 m/s.
(A) The magnitude of force associated with this single step for a person is 1855 N.
(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Given data:
The height of fall is, h = 4.77 cm = 0.0477 m.
The magnitude of downward velocity is, v' = 1.0 m/s.
The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].
The mass of person is, m = 74.2 kg.
The magnitude of force is, F' = 1880 N.
The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].
The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]
Thus, the speed of object during falling is 0.967 m/s.
(A)
Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.
Apply the expression of average force as,
[tex]F =\dfrac{m'v'}{t}[/tex]
Solving as,
[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]
Thus, the magnitude of force associated with this single step for a person is 1855 N.
(B)
The expression for the stress is given as,
[tex]\sigma = \dfrac{F'}{A}[/tex]
Solving as,
[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]
Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
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