QUESTION: If the box is initially at rest at x=0, what is its speed after it has traveled 12.0 m ?
Answer:
The velocity is 7.57/s.
Explanation:
Units aside, [tex]F(x)[/tex] can be written as
[tex]F(x) = 18-0.530x[/tex].
The work done by this force must equal the kinetic energy of the box:
[tex]$\int_0^{12} F(x)dx = \frac{1}{2}mv^2 $[/tex]
[tex]$\int_0^{12} 18-0.530x\: dx = \frac{1}{2}(6.2)v^2 $[/tex]
[tex]$[18(12)-0.530\frac{(12)^2}{2}]- [18(0)-0.530\frac{(0)^2}{2}] = \frac{1}{2}(6.2)v^2 $[/tex]
[tex]$177.84 = \frac{1}{2}(6.2)v^2 $[/tex]
[tex]\boxed{v= 7.57m/s}[/tex]
which is the velocity of the box after it has traveled 12 meters.
