contestada

A steel rod of 0.5 cm diameter and 10 m length is stretched 12 cm. Young’s modulus for this steel is 21 kN/cm2. How much work, in joules, is required to stretch this rod?

Respuesta :

whitneytr12

Answer:

[tex]W=3.024[J][/tex]

Explanation:

The work done by the stretching forces is given by this equation:

[tex]W=\frac{1}{2}\frac{E*A*\Delta L^{2} }{L_{0}}[/tex]  (1)      

Where:

  • W is the work done
  • E is the Young's modulus (E = 21000 N/cm²)
  • A is the cross-sectional area ([tex]A=\pi (D/2)^{2}[\tex]=0.20 cm^{2})
  • ΔL is the amount by which the length of the object changes (ΔL=12 cm)
  • L(0) is the initial length (L(0) = 10 m = 1000 cm)

We just need to put it in the equation (1):

[tex]W=\frac{1}{2}\frac{21000*0.20*12^{2}}{1000}=302.4 [N*cm][/tex]

Therefore the work done is:

[tex]W=3.024[J][/tex]

I hope it helps you!                    

     

Otras preguntas