Question

Max (mass = 15 kg) is hanging from one end of a 13-m long bungee cord that has its other end fixed to a bridge above. The bungee cord has a circular cross section with a diameter of 2.5 cm, and a Young’s modulus of 17 MPa. What is the stress in the bungee cord due to Max’s weight?

Answer 1

The stress in the bungee cord due to the Max's weight is $\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}$ or $\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}$ or $\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}$.

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 $2.5\text{ cm}$.  

The mass m of Max is $15\text{ kg}$.

Concept:

The stress developed in the bungee cord is given by:  

$\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}$                                                                               ... (1)         

Here, $\sigma$ is the stress developed in the bungee cord, $F$is the force due to Max's weight and $A$ is the area of cross-section of the bungee cord.  

The radius of the bungee cord will be the half of its diameter.  

$r=\dfrac{d}{2}$

Substitute $2.5\text{ cm}$ for $d$.  

$r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}$ 

Convert the radius of the bungee cord in meter.  

$r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}$ 

The area of cross-section of the bungee cord is given by:  

$A=\pi r^{2}$  

Substitute $0.0125\text{ m}$ for $r$.  

$A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}$  

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:  

$F=mg$  

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 $9.80\text{ m}/\text{s}^{2}$.  

Substitute $15\text{ kg}$ for m and $9.80\text{ m}/\text{s}^{2}$ for g in above expression.  

$F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}$  

Substitute $147\text{ N}$ for F and $4.980\times10^{-4}\text{ m}^{2}$ for A in equation (1).  

$\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}$

Thus, the stress developed in the bungee cord is $\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}$ or $\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}$ or $\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}$.

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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.