Question

A nylon rope that has a 0.73 cm radius and is 12 meters long is used to lift a 24 kg object straight up. If the object is lifted at constant velocity, how much is the rope stretched by? The Young's modulus of the nylon is 5.0 × 109 N/m2 and use 9.81 m/s2 for the acceleration of free fall. Express your answer in centimeters and round to the nearest hundredth.

Answer 1

Answer:

The rope is stretched by 0.3373 cm

Explanation:

As we know  

Youngs Modulus ($5.0 * 10^9$ N/m2) is Stress divided by strain

Stress = force/Area = (65*9.8)/(pi * (7.3*10^-3)^2)

Starin = change in length/original length of the rope = l/12 meters

Substituting the given values we get  

$5.0 * 10^9$ N/m2 = {($\frac{\frac{24*9.8}{\pi *(7.3*10^{-3})^2} }{\frac{\delta L}{L} }$)

$5.0 * 10^9$ N/m2 = {($\frac{\frac{24*9.8}{\pi *(7.3*10^{-3})^2} }{\frac{\delta L}{12} }$)

$\frac{\delta L}{12} = \frac{\frac{65*9.8}{pi * (7.3*10^-3)^2} }{5.0 * 10^9}$

The rope is stretched by 0.3373 cm