Respuesta :
Answer:
[tex]W=0.74J[/tex]
Explanation:
We use the equation to calculate the work from a point [tex]x_{i}[/tex] to a point [tex]x_{f}[/tex] on a spring:
[tex]W=\frac{1}{2}k(x_{f}^2-x_{i}^2)[/tex]
where [tex]W[/tex] is the work, and [tex]k[/tex] is the spring constant, [tex]k=21N/m[/tex].
For this problem since the spring was first stretched 0.16m, [tex]x_{i}=0.16m[/tex], and then was stretched an aditional 0.15m, so [tex]x_{f}=0.16m+0.15m=0.31m[/tex].
Now we substitute the known values in the equation to find the work:
[tex]W=\frac{1}{2}k(x_{f}^2-x_{i}^2)[/tex]
[tex]W=\frac{1}{2}(21N/m)((0.31m)^2-(0.16m)^2)[/tex]
[tex]W=\frac{1}{2}(21N/m)(0.0961m^2-0.0256m^2)[/tex]
[tex]W=\frac{1}{2}(21N/m)(0.0705m^2)[/tex]
[tex]W=\frac{1}{2}(1.4805Nm)[/tex]
[tex]W=0.74 Nm=0.74J[/tex]
The work that must be done to stretch the spring an aditional 0.15m is 0.74 Joules
