Respuesta :
The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
Further Explanation:
The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.
Given:
The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].
The mass m of Max is [tex]15\text{ kg}[/tex].
Concept:
The stress developed in the bungee cord is given by:
[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex] ... (1)
Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.
The radius of the bungee cord will be the half of its diameter.
[tex]r=\dfrac{d}{2}[/tex]
Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].
[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex]
Convert the radius of the bungee cord in meter.
[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex]
The area of cross-section of the bungee cord is given by:
[tex]A=\pi r^{2}[/tex]
Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].
[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]
The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:
[tex]F=mg[/tex]
Here, m is the mass of Max's body and g is acceleration due to gravity.
Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].
Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.
[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]
Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).
[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]
Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
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Answer Details:
Grade: Senior school
Subject: Physics
Chapter: Stress and Strain
Keywords:
Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.
The stress in the bungee cord due to Max’s weight is 294,000 N/m².
The given parameters;
- mass of Max, m = 15 kg
- diameter of the cord, d = 2.5 cm
- radius of the cord, r = = 1.25 cm = 0.0125 m
- Young's modulus of the rod, E = 17 MPa
The area of the bungee cord is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]
The weight of Max is calculated as follows;
[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]
The stress in the bungee cord due to Max’s weight is calculated as;
[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]
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