Question

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.

Answer 1

Step-by-step explanation:

L = 1 m

A = 0.5 mm² = 0.5 x 10^-6 m²

Y = 2 x 10^11 Pa

F = 1500 N

ΔL = ?

Use the formula for the young's modulus

`Y = (FL)/(A\Delta L)`

`\Delta L = (FL)/(AY)`

`\Delta L = (1500* 1)/(0.5*10^(-6)* 2* 10^(11))`

ΔL = 0.015 m

ΔL = 0.02 m

Answer 2

`\Delta l=0.015m`

Step-by-step explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area
`A=0.5mm^2=0.5* 10^(-6)m^2`

Young's modulus
`\gamma=2* 10^(11)Pa`

Force F = 1500 N

So stress
`=(force)/(area)=(1500)/(0.5* 10^(-6))=3000* 10^(-6)=3* 10^(9)Pa`

We know that young's modulus
`=(stress)/(strain)`

So
`2* 10^(11)=(3* 10^(9))/(strain)`

`strain=1.5* 10^(-2)=0.015m`

Now strain
`=(\Delta l)/(l)`

`0.015=(\Delta l)/(1)`

`\Delta l=0.015m`